For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Continuity & OneSided Limits
 Long ago in this course, we learned about continuous functions. At the time, we lacked the formal ideas to precisely define continuity, so we intuitively defined it as being any of these three equivalent things:
 All the parts of the function are connected;
 The function's graph can be drawn without ever having to lift your pencil from the paper.
 There are no "breaks"/"holes" in the graph.
 Now we have the the formal background to expand on this intuitive definition and define continuity precisely. Formally, a function f(x) is continuous at c if f(c) exists and
This is the same as the above intuitive ideas, because when a limit matches up with its function, it means that the function goes where we "expect"there are no jumps or weird stuff.
lim
f(x) = f(c).  If f(x) is not continuous at c, we say it is discontinuous at c. We call such a location a discontinuity. If f(x) is continuous at every point in an interval (a,b), we say f is continuous on (a,b). If f(x) is continuous for every real number, we simply say that f is a continuous function.
 Notice that the vast majority of the functions we're used to working with are continuous. This is why "normal" functions allow us to simply plug in the value we're approaching. Because we generally have
Furthermore, this also explains why a function that is "weird" in one place still allows us to plug in the value we're approaching if it's not the "weird" value. Because, other than the "weird" value, the rest of the function is continuous, so we can apply the same logic."Normal" ⇒ Continuous ⇒
lim
f(x) = f(c)  If a limit does not exist because the two sides do not agree on a single value, we can consider a onesided limit: the limit of a function as it approaches from only one side.
denotes the limit as x approaches c from the left or negative side. It is the limit as x→ c while x < c.
lim
f(x)
denotes the limit as x approaches c from the right or positive side. It is the limit as x→ c while x > c.
lim
f(x)  For onesided limits, be careful to keep track of which symbol goes with which side; they can be easy to get confused at first. Remember, `−' means look at the `negative' side, while `+' means look at the `positive' side.
 If both sides of a function go to the same value as a location is approached, then the normal limit exists there. Furthermore, if a normal limit exists, both sides must go to the same value as they approach it.
Similarly, if both onesided limits do not go to the same value (or one side doesn't exist), then a normal limit does not exist there.
lim
f(x) = L =
lim
f(x) ⇔
lim
f(x) = L
lim
f(x) ≠
lim
f(x) ⇒
lim
f(x) does not exist  The above idea allows us to find the limits of piecewise functions at "breakover" points. (If it's not a breakover, we can usually just plug the location in, as discussed in the previous lesson.) We find the limit of a piecewise function by checking if the left and rightside limits agree with each other.
 If they agree, then the limit exists and equals them.
 If they do not agree, then the limit does not exist.
Continuity & OneSided Limits

 


 The limits that we see in the problem are onesided limits. A onesided limit works basically the same as a normal limit, except it only considers one side of the function. Here is how we should interpret the extra symbol of − or + after the location the limit is approaching:
denotes the limit as x approaches c from the left (negative) side. It is the limit as x→ c while x < c.
lim
x→ c^{−}f(x)
denotes the limit as x approaches c from the right (positive) side. It is the limit as x→ c while x > c.
lim
x→ c^{+}f(x)  Thus, for this problem, we are looking at the limit as f(x) approaches 3 from the left (x → 3^{−}) and the limit as f(x) approaches 3 from the right (x→ 3^{+}). Notice that, although f(x) is a piecewise function, the only place that is not continuous is at x=3. In other words, while f(x) is "weird" at the "breakover" of x=3, it behaves normally everywhere else. Since the function tells us how it behaves on each side of the breakover and these function pieces are continuous ("normal"), we can use them to find the left and right side limits.
 Use the appropriate piece from the piecewise function to find the value as it approaches from each side. x→ 3^{−}, so use the function piece for x ≤ 3:
lim
x → 3^{−}f(x) = (3) +4
x→ 3^{+}, so use the function piece for x > 3:= 7
lim
x → 3^{+}f(x) = 2(3) − 8 = 6−8 = −2

 

 


 The limits that we see in the problem are onesided limits. A onesided limit works basically the same as a normal limit, except it only considers one side of the function. Here is how we should interpret the extra symbol of − or + after the location the limit is approaching:
denotes the limit as x approaches c from the left (negative) side. It is the limit as x→ c while x < c.
lim
x→ c^{−}f(x)
denotes the limit as x approaches c from the right (positive) side. It is the limit as x→ c while x > c.
lim
x→ c^{+}f(x)  Thus, for this problem, we are looking at the limit as g(x) approaches −4 from the left (x → −4^{−}) and the limit as g(x) approaches −4 from the right (x→ −4^{+}). Notice that, although g(x) is a piecewise function, the only places that are not continuous are at x=−4 and x=−3. In other words, while g(x) is "weird" at those "breakovers", it behaves normally everywhere else. Since the function tells us how it behaves on each portion and these function pieces are continuous ("normal"), we can use them to find the left and right side limits.
 Use the appropriate piece from the piecewise function to find the value as it approaches from each side. x→ −4^{−}, so use the function piece for x < −4:
lim
x → −4^{−}g(x) = 3(−4) +1 = −12+1
x→ −4^{+}, so use the function piece for −4 ≤ x <−3:= −11
lim
x → −4^{+}g(x) = (−4)^{2} −2 = 16−2
[Remark: Notice that for x→ −4^{+}, we use the function piece based on −4 ≤ x <−3 but not the piece for −3 ≤ x. This is because the right side limit is based on the part of the function that is just slightly to the right of −4. While the piece for −3 ≤ x is to the right of −4, it is not adjacent to the location, so we do not use it. On the other hand, −4 ≤ x <−3 is "touching" the limit location on the right side, so we use that piece of the function.]= 14
 Begin by noticing that the function x^{3}−x^{2}+3 is continuous ("normal"). It is connected everywhere and does not behave "weirdly" at any location. This means that the normal limit of the function exists everywhere and equals the value of the function:
lim
x→ cf(x) = f(c)  From the lesson, we learned that if the normal ("twosided") limit of a function exists at some location, then both onesided limits exist at the location and are equal to each other.
lim
x→ cf(x) = L ⇔
lim
x→ c^{−}f(x) = L =
lim
x→ c^{+}f(x)  Therefore to find both of the onesided limits in the problem, we just need to find the limit of the function for x→ 2. Furthermore, since the function is continuous ("normal") we can just plug in the limit location of c=2 to find the value of the limit:
lim
x → 2x^{3}−x^{2}+3 = (2)^{3} − (2)^{2} + 3 = 8 − 4 + 3 = 7

 


 Start off by noticing a small but very important fact: while the piecewise function has its "breakover" location at x=1, the onesided limits are x → −1^{−} and x → −1^{+}, so they are about the location −1. This means that, while f(x) does have two different pieces, we are only interested in one of those pieces. Since the breakover location is not the location that the limit is focused on, we do not have to care about the function piece that does not contain the limit location.
 Since we're only interested in the function piece that contains the limit location of −1, we only need to work with 2x+5. Next, notice that 2x+5 is continuous ("normal"). This means that the normal limit of the function exists and equals the value of the function:
Furthermore, because the normal ("twosided") limit exists then we know both onesided limits exist at the location and must equal each other.
lim
x→ cf(x) = f(c)
lim
x→ cf(x) = L ⇔
lim
x→ c^{−}f(x) = L =
lim
x→ c^{+}f(x)  Therefore to find both of the onesided limits in the problem, we just need to find the limit of the function for x→ −1. Also, since the function is continuous ("normal") at the location since it is not a piecewise breakover, we can just plug in the limit location of c=−1 to find the value of the limit:
lim
x → −1g(x) = 2(−1)+5 = −2+5 = 3
 A function is continuous at c when f(c) exists and
A discontinuity is the opposite of the above. A discontinuity can happen because the function does not exist at the location, the limit does not exist at the location, or the limit does not match the value of the function at the location.
lim
x→ cf(x) = f(c)  To do this, figure out all the locations where f(x) is not defined. In other words, find where the function "breaks". Notice that [(x−2)/(x^{2}−x−2)] can "break" because of dividing by 0. Factor the denominator to find where this can happen:
Thus f(x) is not defined at x=−1 or x=2, so there is a discontinuity at each of those locations.x−2 x^{2}−x−2= x−2 (x+1)(x−2)
[Remark: Notice that, although we can simplify the function as
this does not get rid of the function "breaking" at x=2. While we can simplify functions by canceling factors to find limits, this does not change the domain of the original function. The domain is based on the function before simplification, so x=2 is not a part of the domain even if we can later "fix" the hole with simplification.]x−2 x^{2}−x−2= x−2 (x+1)(x−2)⇒ 1 x+1,
 A function is continuous at c when f(c) exists and
Notice that f(x) can "break": taking the square root of a negative number is not defined, so the function will not exist whenever that happens.
lim
x→ cf(x) = f(c)  Figure out when a negative value would occur beneath the square root. There are two possibilities for the absolute value x: either x is positive (or 0) and it has no effect, or x is negative and it flips the sign. Deal with each of these cases separately. If x is positive, then the value under the square root is negative when
Therefore the value under the square root is negative when 0 ≤ x < 1. If x is negative, then the value under the square root is negative whenx−1 < 0 ⇒ x < 1
Therefore the value under the square root is negative when −1 < x ≤ 0. Since either one of these conditions will cause the value under the square root to be negative, that means the function is not defined ("breaks") when we have −1 < x < 1.−x−1 < 0 ⇒ −1 < x  Notice that for any other value, the function will be defined. That is, for any x where x ≤ −1 or x ≥ 1, the function will work "normally". How does this happen? Because for x ≤ −1, the function behaves as if it were just √{−x−1}, while for x ≥ 1, the function behaves as if it were just √{x−1}. With this in mind, let us graph the function to better understand it:
 Based on this graph and our understanding of how the function works, we see that for any x where x < −1 or x > 1, the function works "normally": the function exists at the location, the limit exists at the location, and the two match each other. Thus for x < −1 and x > 1, the function is continuous. From earlier, we saw that the function does not exist for −1 < x < 1, so it is not continuous at any of those values. This leaves us with two points to investigate: what happens at x=−1 and x=1? Clearly, from the graph and by using the function, we see that these locations do not "break" the function and f(−1) = f(1) = 0. However, for a point to be continuous, it must exist and have a matching limit at that location. Let us now figure out if limits exist at these locations. For ease, let's just consider x=1 at first. Notice that while the function exists for x ≥ 1, for x < 1 the function does not exist. But to have a limit, the function must be approaching the same value from both sides. We see that while the function exists to the right of x=1, there is nothing adjacent to x=1 on its left. Thus, the limit as x → 1 does not exist. (Notice that the rightside limit of x→ 1^{+} would exist, but that's not enough: continuity is based on the normal, twosided limit.) We see that the exact same thing happens for x=−1, but in the opposite way. Thus the limit as x→ −1 does not exist either. Therefore, because the limits do not exist at these locations, they cannot be continuous. Combining that with our earlier observations, we now see that f(x) is discontinuous for −1 ≤ x ≤ 1. Thus the continuous points must be the opposite of the above, which is x < −1 and x > 1. In interval notation, we write that as (−∞, −1) and (1, ∞).

 


 Begin by noticing that each piece of the piecewise function is "normal". That is, each piece of the function is continuous on its own. The issue comes at the "breakover" point where we switch from the first function piece to the second function piece. We need to know whether or not these two function pieces are headed towards the same location. In other words, we need to know if the left and rightside limits of x→ 2^{−} and x→ 2^{+} match up with each other.
 One way we could do this problem is by making a highquality graph to see if the function pieces "connect" with each other at the breakover location of x=2. However, making a good graph takes a lot of effort and it is impossible to make an absolutely perfect graph. Instead, an easier way is to just figure out what the values being approached by the left and right sides are. If the limit as x→ 2^{−} matches the limit as x→ 2^{+}, then the pieces "connect" and the function is continuous. If they do not, then it is not continuous.
 Find the values being approached by each side. For x→ 2^{−}, we use the function piece based on x < 2 because we're considering the values to the left of x=2:
For x→ 2^{+}, we use the function piece based on x ≥ 2 because we're considering the values to the right of x=2:
lim
x → 2^{−}f(x) = 3(2) −7 = 6−7 = −1
Because the left and right sides approach the same value, we see that the normal, twosided limit exists there. This means that the function "connects" its two pieces, so the function as a whole is continuous.
lim
x → 2^{+}f(x) = −2(2) + 3 = −4 + 3 = −1

 


 Begin by noticing that each piece of the piecewise function is "normal". That is, each piece of the function is continuous on its own. The issue comes at the "breakover" point where we switch from the first function piece to the second function piece. We need to know whether or not these two function pieces are headed towards the same location. In other words, we need to know if the left and rightside limits of x→ −3^{−} and x→ −3^{+} match up with each other.
 One way we could do this problem is by making a highquality graph to see if the function pieces "connect" with each other at the breakover location of x=−3. However, making a good graph takes a lot of effort and it is impossible to make an absolutely perfect graph. Instead, an easier way is to just figure out what the values being approached by the left and right sides are. If the limit as x→ −3^{−} matches the limit as x→ −3^{+}, then the pieces "connect" and the function is continuous. If they do not, then it is not continuous.
 Find the values being approached by each side. For x→ −3^{−}, we use the function piece based on x ≤ −3 because we're considering the values to the left of x=−3:
For x→ −3^{+}, we use the function piece based on x >−3 because we're considering the values to the right of x=−3:
lim
x → −3^{−}g(x) = −(−3)^{2}−4(−3)+2 = −(9) +12 +2 = 5
Because the left and right sides do not approach the same value (5 ≠ −7), we see that the normal, twosided limit does not exist there. This means that the function fails to "connect" its two pieces, so the function as a whole is not continuous. [Remark: Notice that each piece on its own is still continuous, the only issue is the breakover point. This means that g(x) is continuous on (−∞, −3) and (−3, ∞). It's just that since g(x) is not continuous at x=−3, it is not continuous everywhere, so we do not call it a continuous function.]
lim
x → −3^{+}g(x) = 2(−3) − 1 = −6−1 = −7

 Begin by noticing that each piece of the piecewise function is "normal". That is, each piece of the function is continuous on its own. [While the value we choose for k will affect the second piece of the function, it will wind up being "normal"/continuous no matter what value we choose.] The issue comes at the "breakover" point where we switch from the first function piece to the second function piece. For the function to be continuous, we need these two pieces to "connect" together. To help us understand what's going on, we can imagine the problem like this: the first function piece is fixed in its location, but we are allowed to adjust the "height" of the second function piece (because we can change k, the yintercept). We can visualize this with the below graph: the red part on the left is the first piece, while the other colored pieces represent some possible heights to place the second piece at by adjusting k.
 What we need to do is figure out the value for k that will "connect" these two pieces together (in the graph above, this would be how the red and blue pieces connect at the breakover). Formally, this means we need the left and rightside limits to give the same value. If these sides match up to the same value, then the pieces "connect" and we will have a continuous function. Therefore we need to figure out what value for k will make the below equation true:
lim
x→ 1^{−}f(x) =
lim
x→ 1^{+}f(x)  Notice that the limit as x → 1^{−} is based on the left side, so it uses the function piece where x < 1. Similarly, x→ 1^{+} is based on the right side, so it uses the piece where x ≥ 1:
lim
x→ 1^{−}f(x) =
lim
x→ 1^{+}f(x) −2(1) + 4 = (1) + k
We can now solve the resulting equation for k to get that k=1. Therefore, to make the two sides "connect" and cause the function to be continuous, we must have k=1.2 = 1+k

 Begin by noticing that each piece of the piecewise function is "normal". That is, each piece of the function is continuous on its own. [While the value we choose for c will affect where each function piece is used, both pieces will still be "normal"/continuous no matter what we choose for c.] The issue comes at the "breakover" location (which we are choosing since we are figuring out c). For the function to be continuous, we must choose c such that the two pieces will "connect" at the breakover. To help us understand what's going on, we can imagine both function pieces being graphed as if there were no conditions on either of them. Our job is then to choose the value of c that matches up to where they "connect". This will give us a breakover point that is continuous.
 More formally, we are trying to make the left and rightside limits match up to each other at the location of x=c. While we do not know the value for c yet, we do know that if they match up, we will have
Therefore, if we can figure out a value for c that makes the above equation true, we will have a continuous function.
lim
x→ c^{−}g(x) =
lim
x→ c^{+}g(x)  Even if we don't know what the value of c is yet, we can still consider how the limit interacts with the function. Notice that if we are looking at the limit as x→ c^{−}, then we are looking from the leftside, so x < c. Similarly, x→ c^{+} uses the rightside, so x > c. Using the appropriate pieces from our piecewise function, we have
c+8 = 3(c) 2\] We can now solve this equation to find c:
lim
x→ c^{−}g(x) =
lim
x→ c^{+}g(x) c+8 = −3c −2 4c = −10
Therefore, if we set c = −[5/2], then the breakover will be in the right location to cause the two pieces of the function to "connect", and therefore the function will be continuous.c = − 5 2
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Continuity & OneSided Limits
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 Motivating Example
 Continuity  Idea
 Continuous Function
 All Parts of Function Are Connected
 Function's Graph Can Be Drawn Without Lifting Pencil
 There Are No Breaks or Holes in Graph
 Continuity  Idea, cont.
 We Can Interpret the Break in the Continuity of f(x) as an Issue With the Function 'Jumping'
 Continuity  Definition
 A Break in Continuity is Caused By the Limit Not Matching Up With What the Function Does
 Discontinuous
 Discontinuity
 Continuity and 'Normal' Functions
 Return of the Motivating Example
 OneSided Limit  Definition
 OneSided Limit  Example
 Normal Limits and OneSided Limits
 Limits of Piecewise Functions
 'Breakover' Points
 We Find the Limit of a Piecewise Function By Checking If the Left and Right Side Limits Agree With Each Other
 Example 1
 Example 2
 Example 3
 Example 4
 Intro 0:00
 Introduction 0:06
 Motivating Example 0:56
 Continuity  Idea 2:14
 Continuous Function
 All Parts of Function Are Connected
 Function's Graph Can Be Drawn Without Lifting Pencil
 There Are No Breaks or Holes in Graph
 Continuity  Idea, cont.
 We Can Interpret the Break in the Continuity of f(x) as an Issue With the Function 'Jumping'
 Continuity  Definition 5:16
 A Break in Continuity is Caused By the Limit Not Matching Up With What the Function Does
 Discontinuous
 Discontinuity
 Continuity and 'Normal' Functions 6:48
 Return of the Motivating Example 8:14
 OneSided Limit
 OneSided Limit  Definition 9:16
 Only Considers One Side
 Be Careful to Keep Track of Which Symbol Goes With Which Side
 OneSided Limit  Example 10:50
 There Does Not Necessarily Need to Be a Connection Between Left or Right Side Limits
 Normal Limits and OneSided Limits 12:08
 Limits of Piecewise Functions 14:12
 'Breakover' Points
 We Find the Limit of a Piecewise Function By Checking If the Left and Right Side Limits Agree With Each Other
 Example 1 16:40
 Example 2 18:54
 Example 3 22:00
 Example 4 26:36
Precalculus with Limits Online Course
Transcription: Continuity & OneSided Limits
Hiwelcome back to Educator.com.0000
Today, we are going to talk about continuity and onesided limits.0002
By now, we have a pretty good intuitive understanding of how limits work, and a reasonable sense of how we can evaluate them precisely.0005
However, we haven't really worked with piecewise functions yet.0011
In this lesson, we will not only learn how to apply limits to piecewise functions, but in doing so,0015
we will learn about some new ideas involving limits.0019
Our exploration will deepen our understanding of what it means for a function to be continuous, and also reveal the idea of a onesided limit.0022
And if you don't quite remember how piecewise functions work, or you find them confusing,0028
make sure that you check out the lesson on piecewise functions that we did a long, long, long time ago,0032
almost 40 lessons backmaybe even more; we did a lesson on piecewise functions.0037
So, if you don't have an understanding of how piecewise functions work, you probably want to check that out first,0042
because the idea of piecewise functions will end up showing up a lot when we are working with limits, especially in this lesson.0045
All right, let's go: first we will set up a motivating example, so that we can have some realizations come out of this motivating example.0050
This piecewise function example will help lead us to some ideas.0058
We have 1 when x is less than 0; that is the portion right here.0062
We have 0 when x equals 0 (that is the portion right there); and we have positive 1 when x is greater than 0.0067
And notice how there are holes at actually equal to 0 on the top and actually equal to 0 on the bottom,0075
because it jumps to the middle portion, right here, when it is actually at x = 0.0084
OK, notice that, while f(0) exists (we get a valuewe get f(0) = 0), the limit as x goes to 0 does not exist.0091
As we show up from the two different sides, as we come in, they don't agree with each other.0100
So, the point in the middle, f(0), does exist; but the limit does not exist.0107
The issue is not so much that f(x) is a weird thing; it is not particularly weird.0110
It is more an issue of f(x) all of a sudden doing this jump, where it is doing this giant leaping around.0116
It is jumping from one place to another...wait a second...jumping?0122
That sounds familiarthe idea of moving around suddenly...breaking the graph...0127
Long, long ago, in this course, when we first learned about piecewise functions, about 40 lessons ago, we mentioned the idea of a continuous function.0132
At the time, we lacked the formal ideas to precisely define continuity.0142
So, we intuitively defined it as being any of these three equivalent things:0146
all of the parts of the function are connectedwe have some function, and everything in the function connects together;0150
alternatively, we can think of this as that the function can be drawn without ever having to lift your pencil from the paper,0156
because if you have to lift your pencil, if we draw part of it, but then you have to stop here,0164
and lift your pencil to go somewhere else to keep going, well, we have this break here;0169
and because of that break, it means that it is not continuous.0174
And finally, there are no breaks or holes in the graph.0177
We see a break pretty clearly in this, where we go one way, and then all of a sudden it switches to something else.0180
But we can also it just having a hole as being the same issue, where it goes up here,0186
but then for some reason this point is not defined, and then it keeps going.0191
It is continuous, except for the part where it has this little break in the middle, which breaks continuity.0194
At this point, we actually now have the formal background to expand on this intuitive definition of it not being connected,0200
it not being continuous, and actually define continuity in a precise term, using limits.0207
To help us see this, we will compare a piecewise example with a continuous function.0213
Here is our piecewise example that we have been working with so far, and here is our continuous function friend0217
for any time we need to pull out a random function, g(x) = x^{2}, a parabola.0224
Over here in our piecewise break, we can interpret the break in continuity, this part where it jumps, as it jumping.0230
The issue with it being continuous isn't that something breaks in the function.0237
The function is perfectly defined at every number; any number that you put in...it knows what to go out to.0242
But the issue is that it jumps all of a sudden; that is, at x = 0, it doesn't go where we expect; it does not go to the place that we expect.0248
Either direction we come at this thing, no matter how we look at it, we can't build an expectation.0256
We can't build an expectation for where the next place is, because there is no agreement; we can't agree on where we are going.0261
Formally, when we can't have an expectationthere is no agreement on where we are going0267
as we come from the two sidesthat means that there is no limit as it goes to x = 0.0271
As we get towards x = 0, limit as x goes to 0 does not exist, because there is no agreement on the value that they are going to.0276
Those two sides are going to two totally different values.0283
On the other hand, in g(x), there are no jumps; the nice thing about g(x), the nice thing about the parabola,0285
is that the function goes where we expect; it always ends up showing up where we expect.0291
Every point is exactly where we expect that point to be.0297
That is to say, every point always has a limit, and that point matches with that limit.0300
All of g(x) is exactly where we expect it to be.0306
The limits and the points in the function are exactly the same thing.0308
This is the idea that we use for continuity: we see that a break in continuity, the break in a function,0312
is caused by the limit...our expectation not matching up with what the function does.0318
Formally, we say that a function f(x) is continuous at some location c if f(c) exists and the limit as x goes to c of f(x) equals f(c).0323
In other words, the function gives the value we expect when we evaluate it at c.0335
The value that we expect to come out as we get close to c is exactly what the value ends up being.0341
We expect some value, and then that value turns out to actually be what it is.0346
That idea of the limit matching what the function actually is there is what it means for something to be continuous at a point.0350
Expanding on this idea, we have more vocabularyjust ways to talk about this thing.0357
If f(x) is not continuous at c, we say that it is discontinuous at c; the opposite of continuous is discontinuous.0361
If we want to talk about some specific location where it breaks from being continuous, we say that it is a discontinuity.0369
A single discontinuous point is a discontinuity.0375
If f(x) is continuous at every point in an interval, over an entire interval0380
it is always continuous throughout that intervalwe say f is continuous on (a,b), on some interval.0385
If f(x) is continuous for every real numberit is always continuouswe simply say that f is a continuous function.0391
The function is continuous if it always ends up working out continuously; great.0399
Notice that the vast majority of the functions that we are used to working with are continuous functions.0405
When we think of functions, we usually think of these nice, smooth, curved things that all fit together quite cleanly.0409
So, when we think "normal," what we are really thinking is "continuous."0416
This is why normal functions allow us to simply plug in the value we are approaching0420
because we generally have...a "normal" function, normally, in our minds, means that it is a continuous function.0424
And by definition, a continuous function is one where the limit as x goes to c of f(x) is the same thing as f at c0430
that we can just plug in the c for our x, and we will end up getting the same thing as if we looked at the limit.0437
The expectation is where we actually come out to be.0442
Furthermore, this idea also explains why a function that is weird in one place still allows us to plug in the value we are approaching if it is not the weird value.0445
That is because, if it is not the weird value, then that means that it is normal in that area;0453
or more accurately, it is continuous in that place, if it is not the weird place.0458
And because it is continuous in that place, that means we still have the same thing: that the limit as x goes to c of f(x) will match with f(c).0462
That means that we can find out what the limit is by simply plugging in c into our function.0470
So, the same logic is what we can apply here.0474
And we end up seeing that, as long as it is normal in this region around c, as long as there is0477
some tiny little neighborhood around c that behaves normally (which is to say continuously),0481
we can just plug c in, and we will end up being able to figure out, "Oh, that is what the limit comes out to be."0486
Let us return to our motivating example, now that we have the notion of continuity learned.0492
Back to our motivating example: we have 1 when x is less than 0, 0 when x equals 0, and 1 when x is greater than 0.0497
On the one hand, as we approach 0, we totally don't have a limit.0503
As x goes to 0, it does not produce a limit, because they don't agree.0508
If they don't agree, it is not a limit; both sides have to agree.0512
However, if we were to imagine only approaching from a single side, all of a sudden, it starts to work; we could find limits.0516
This realization leads us to the creation of a new type of limit, the onesided limit.0527
If we approach only from the negative side, it is clear that we are headed to 1.0531
If we were to approach only from the positive side, it is clear that we would be headed to positive 1.0535
So, we can have this idea of a onesided limit, where we just look at approaching from one side or the other.0540
We don't worry about what the other side is doing; we just worry about what this side is doing;0545
where is it headed towards, as we stay coming from this one direction?0550
A onesided limit basically works just the same as a normal limit, except that it only considers one side; it is only looking at one side.0555
The limit as x goes to c with a negative symbol in the top rightour c that we are going to, and then a negative symbol in the top right0563
denotes the limit as x approaches c from the left, or the negative side, equivalently.0570
It is the limit as x goes to c, while x is less than c.0576
So, as x remains less than cthat is, as we come from the left side, the negative side, that is the limit that we end up getting:0579
limit as x goes to c, with that little negative sign in the corner.0587
Similarly, if we have limit as x goes to c with a positive sign in the corner, that denotes the limit as x approaches c from the right, or positive, side.0590
As we come from the right side, the positive sidethat is going to be all of the x's where x is greater than c.0598
So, as our x stays above c, what are we going towards?0604
Be careful to keep track of which symbol goes with which side.0608
They are easy to get confused at first, so be careful about this.0612
Remember: the negative symbol means to look at the negative side; and positive means to look at the positive side.0616
If we say that x goes to c with a negative symbol in that top right corner,0626
that means that we are saying to look as we come from the left side, the negative side.0631
If it is x goes to c with a positive sign in that top right corner, we are saying,0636
"Look at what happens as we come from the right side, as we come from the positive side."0640
All right, let's see this idea applied to our example.0646
We have 1 when x is less than 0; so as we go in from the left side, limit as x goes to 0 with a negative sign,0650
we get 1, because that is the value that we appear to be approaching.0658
If we come from the positive side, the limit as x goes to 0 from the positive side, we end up approaching positive 1.0662
If we just simply evaluate what f at 0 is, we get just simply 0.0670
Now, notice: there is not necessarily any connection between the left and right side or the actual value of what the function is.0677
1 is totally different than 0, which is totally different than positive 1.0688
The limits, the two side limits, don't agree, and the actual value that comes out of the function doesn't agree with either of them.0692
So, each thing can behave totally independently.0698
However, one connection is that the two different side limits don't agree.0700
And because they don't agree, that means that there can't be a normal limit, the limit of it coming from both sides.0705
For there to be a normal limit, they have to match up; if they don't match up, then that means that there is no normal limit.0710
So, matching up: if the left and right side, the two sides, match up together, that means that we do have a normal limit.0716
If, on the other hand, they don't match up, then that means that we don't have a normal limit.0723
So, normal limits and onesided limits: if both sides of a function go to the same value0727
as some location is approached, then the normal limit exists there.0735
Furthermore, if a normal limit exists, then both sides must go to the same value as they approach that location.0739
Symbolically, we can write this as the leftside limit equals l, and the rightside limit equals l,0748
implies that the normal limit is equal to l, as well.0756
If our left side goes to the same value as our right side, then that means that the sides match up.0760
So, we have a normal limit; and because they both went to l, then it comes out as an l limit, a limit going to l.0768
Similarly, if we went the other direction, if we know that there is a limit here at l;0774
then that means that, if we went off to the left side, or we went off to the right side,0778
we are going to end up having it be l in both of those directions, as well.0782
So, if the normal limit exists, then we know that the leftside limit and the rightside limit,0787
the negative limit and the positive limit, have to both be equal to the same thing.0795
They have to be equal to this same l.0799
All right, continuing with the same idea: if both onesided limits do not go to the same value0801
that is, they don't match up, or one of them simply doesn't exist at allthen a normal limit does not exist there.0808
That is to say, if the leftside limit is not the same as the rightside limit, then that means that the normal limit does not exist.0814
And if the normal limit does not exist, that tells us that one of them has to not exist, or that the sides don't match up to each other.0825
So, there is this connection between normal limits and onesided limits.0834
If the onesided limits meet up together, there is a normal limit at that place.0837
If there is a normal limit at that place, then that means that the onesided limits must match up together.0840
And the same thing happens if the onesided limits don't match up: then that means that there is no normal limit.0844
If there is no normal limit, then we can't have matching up; great.0848
All right, limits of piecewise functions: we are now ready to actually talk about how to apply all of this stuff to a piecewise function.0851
It is this idea about "Do they match up? Do they not match up?" that lets us find the limits of a piecewise function at a breakover point.0857
A breakover point is switching from one function to another function.0867
We are going here, and then, all of a sudden, we come out as some other thing.0871
Let's see what we are talking about here: it is some function here, and then all of a sudden, it swaps over to being some other function, like this.0874
So, I am talking about this breakover place being where it switches between the two.0884
If it is a breakover point, and we want to evaluate this, well, we can say, "Where were you going from the left side?0892
Where were you going from the right side?" and we can ask questions about each of the onesided limits and how they relate to each other.0897
And if it is not a breakover point, then we don't have to worry about that, because that normally means0902
that we are just in the middle of a normal, nice, continuous part of a piecewise function.0907
And that means that we can usually just plug the location, as we discussed in the previous lesson, and as we sort of discussed earlier in this lesson.0912
If you are in the middle of a nice continuous chunk of a function, you just plug in the location that you are going to,0920
because "continuous" means that the limit matches up with what it evaluates to be there.0926
So, if we do have a piecewise function, though, and we are looking at a breakover,0931
where we switch from one track to another track, then we can find the limit of a piecewise function0934
by checking if left and right side agree; if the left side and the right side agree with each other,0943
then that means that we have a limit there; so if they agree, the limit exists.0952
And it is equal to what they both are, because they have to match up to the same value.0957
If they do not agree (one goes here and the other one goes here), then there is a jump between them,0961
which means that there can't be a normal limit, because we have to agree from both sides; then, the limit does not exist.0966
Since piecewise functions are usually made up of fairly normal functions0972
(you don't really normally see any very weird functions when we are dealing with piecewise functions),0975
since each side is normally fairly normal, it is normally pretty easy to find the left and right side limits,0979
when we want to compare them, because we can just say, "OK, if we plug in for the left side over here,0984
and plug in for the right side over here, we are able to figure out that that is what the leftside limit is;0988
that is what the rightside limit has to be; now decide whether or not they actually match up."0992
So, you can normally pretty easily figure out what they are.0996
All right, let's see some examples: the first example: Find both of the onesided limits below0998
for f(x) = 3x + 3 when x < 2, and x^{2} + 2 when x ≥ 2.1004
First, let's just see a quick sketch to get some idea of how this thing works.1012
3x + 3 when x is less than 2: we are going to end up being at a fairly steep incline like this.1017
And it pops out of existence here, because now we hit x < 2.1029
And then, we switch over to some x^{2} + 2; x is greater than or equal to 2;1034
so, at x^{2} + 2, we would pop up here; and then we would continue on our way,1039
with whatever the parabola is, already in motion.1043
All right, that is what we are seeing from this piecewise function.1047
When it says "the limits below for x going to 2 from the negative side," well, if it is from the negative side,1050
we are looking for what is happening as we come in here.1056
So, if we are looking at what is coming in here, then that is x less than 2, which means we are just concerned about 3x + 3.1059
So, if that is the case, then that means limit as x goes to 2 from the left side is just...1067
if we plugged in 3x + 3, and we plug in 2 (it is 2 from the left side; we plug in 2), 3(2) + 3 is 6 + 3, so we get 3 there.1072
Over here, if we want to talk about the limit as x goes to 2 from the positive side,1092
then what we are concerned about is as we come in from this side; as we come in from this side,1096
we belong to x being greater than or equal to 2; we don't actually have to worry about the x = 2,1101
because they are both limits, so it is only the journey towards that location.1106
We use x^{2} + 2 if we want to talk about what it is going to be.1110
x^{2} + 2 was nice and normal until that flipover; so we can just plug into that.1113
x^{2} + 2 means it is going to be 2^{2} + 2, 4 + 2, or 6.1117
And that is what we get for our two different sides for the limits here.1127
Notice that they don't match up.1130
All right, the second example: Evaluate the limit as x goes to 3 from the positive side of √(x  3) + 4.1132
First, let's just quickly draw a quick sketch, so that we can see what is going on here.1137
So, √(x  3) + 4: where would that start?1142
Well, the first value that would make sense in there...if we plug in anything less than 3,1148
we are going to be taking the square root of a negative number, so it can't be any less than 3.1152
So, the very first point we could plot is at +3; and we would have 4 come out of it.1155
And then, it would curve out like a square root function, like this.1160
OK, if we are concerned about x going to 3 from the positive side, then what we care about is what happens on our way there.1163
On our way there, it ends up just behaving like √(x  3) + 4.1170
It is totally normal, up until the moment where it stops existing completely, when we try to take square roots of negatives.1174
The limit as x goes to 3 from the positive side, of √(x  3) + 4...1180
well, since it behaves totally normally up until the point where it stops existing,1189
but we don't care about that side, because we are on the existing side, the positive sidewe can just plug in our value for 3.1193
√(3  3) + 4 is √0 + 4, or positive 4; and that totally makes sense.1202
The point that we expect to go to is this one right here, where it starts.1211
There are no weird jumps; there is nothing weird going on as we go in.1216
As we come in from the positive side, it is very clearly headed towards 4, so it makes total sense that it has a limit.1219
Now, what about explaining why the limit as x goes to 3 of √(x  3) + 4 does not exist?1230
Well, the real thing here is: does it come into the same thing from both sides?no, because there is nothing over here at all.1236
If we try to come in from the left side, what is going on?1243
We have no idea what is going on; that is why it does not exist.1245
If we want to get even more formal, we can say that, for the limit as x goes to 3 to exist,1248
that must mean that the limit as x goes to 3 from the positive side is the same thing as limit as x goes to 3 from the negative side.1256
But clearly, the limit as x goes to 3 from the negative side of f(x) (this being f(x) in this case) does not exist.1267
You can't say that it is something that we are headed towards, because it just doesn't exist in that area.1279
If it doesn't exist, as we try to come in from the left side, there is no limit to say that it is going to.1284
So, if the right side simply does not exist, then that means that the normal limit can't exist, because we only have onehalf of a normal limit.1291
Therefore, the limit as x goes to 3 does not exist; and that is why we see that.1298
But more informally, it is just the fact that we can't see it coming to the same thing from both sides.1309
We have to have it from both sides; so if one side simply just isn't there, then it doesn't exist.1313
The third example: Determine if the limit exists; if so, evaluate it.1319
We have a piecewise function here: x^{3} + 4 for x < 1; 7 when x = 1; and (x  1)^{2}  1 when x > 1.1324
So, our first question is, "Where are we headed towards?"1336
We are headed to x going to 1; so we don't just have it being in a nice, convenient, normal section.1338
We are going exactly in the breakover; all right, well, if that is the case, how do we check to make sure it exists?1345
Well, it only exists if from the right side and the left side it meets up to the same value.1352
The two different sided limits have to agree with each other.1358
If that is the case, what we are looking for is the limit as x goes to 1 of f(x); it is going to be based on the limit,1362
as x goes to 1 from the negative side (and don't get confused about the 1 and then the negative;1374
the negative in the top right tells us that we are coming from the negative side; the negative in the normal place1381
just means that it is a negative number), is going to be the same thing as the limit as x goes to 1 from the positive side.1385
If they end up being the same value for the limit, then we end up getting that it does exist as an actual limit, and it is that location.1394
The first thing to notice at this point is...do we care about 7 when x = 1?1402
No, we don't care about it, because it is x going to 1.1408
And remember: it is about the journey, not the destination; it is about where you are headed,1411
but you don't actually care about where you show up, when you are looking at a limit.1415
You just care about where it seemed like you were going to.1418
So, where we actually end up going doesn't matter; we don't have to worry about 7 at all.1420
It is just there as a distraction to get us confused.1425
The only things that we really have to care about are the x^{3} + 4, the part that we are coming from on the left side,1427
x less than 1, and the (x  1)^{2}  1, the part that we are coming from on the positive side, because it is x greater than 1.1433
So, if that is the case, let's work this out; we know that our left side limit will be based off of x^{3} + 4,1442
so that the limit as x goes to 1 from the negative side of f(x) will be the same as if we had simply plugged in 1 for our x here,1452
since we are coming from the left side, and x^{3} + 4 is the way that the thing behaves as long as it is on the left side of 1.1464
So, we plug into there; we have (1)^{3} + 4; 1 cubed is just 1, plus 4...so we get +3 from the left side.1472
Switch to looking at our right side now; the question is, "Do the two sides agree, or do they disagree?"1483
If they go to different places, then the limit does not exist; if they go to the same place, then they do exist, and the limit is where they meet up.1489
So, in this case, we are now looking at...the rightside function is (x  1)^{2}  1.1495
We are looking at the limit, as x comes in from the right side (that is, the positive side), of f(x).1502
Once again, all we are concerned with is the right side of this, and the right side is entirely determined by this function,1510
because how that is how the right side is behaving that whole time, just as the left side was behaving the whole time for x^{3} + 4.1516
So, it is just a question of how that will fit into there, because (x  1)^{2}  1 is a nice continuous function.1521
So, we can just plug into it, because we don't have to worry about anything weird happening.1527
We plug into that, and we have...1 swaps out for our x, minus 1, squared, minus 1; 2 inside of there, squared, minus 1;1531
so that gets us 4  1, which gets us 3; look at that: 3 and 3 check out.1541
So, that means that indeed the limit does exist.1548
Thus, we can combine these two things, and we know that, since the left and rightside limits agreed with each other1551
they both came out to be 3that means that the limit as x approaches 1 from both sides of f(x) is equal to 3.1556
I do want to point out to you, though, that f(x) is not continuous.1567
Why? Because at x = 1, it jumps; so the left side expects to go to 3; the right side expects to go to 3.1570
But when we actually get to 3, it jumps up to 7.1578
So, because it jumps to somewhere else, it is not continuous.1580
The limit exists, but it is not actually a continuous function.1583
It has to have a limit and agree with what that point actually comes out to be, to be continuous.1586
All right, our final one, where we do actually talk about continuity: Determine the value of a that makes f(x) continuous.1592
What does it mean to be continuous? (I will write that out as cts, just because I am lazy).1600
To be continuous: that means that the limit, as x goes to c, of f(x), is equal to f(c).1605
Remember: we talked about this as the expectation for the function being the same thing as what we actually get out of the function.1612
That is what it means for something to be continuous.1618
The expectation, the limit, is the same thing as what we actually get out, f(c).1621
So, to be continuous, the limit as x goes to c of f(x) must be equal to f(c).1625
However, in this case, we don't just have to worry about limits; we also have to worry about the fact that this is a piecewise function.1629
Since it is a piecewise function, we need to make sure that both of the pieces end up agreeing.1635
The first question is, "Does the limit exist?"1641
Well, our question herethe function will be continuous if the limit as x goes to 1 of f(x) is equal to f at 1.1645
All right, that is the case; so what is f at 1?1658
We will come back to the left side and right side in just a second.1661
So, what is f at 1? Well, f at 1 is equal to 5 minus...1663
Oh, which one do we have to use? We use x less than or equal to 1, so we are using this right here.1668
So, 5  2: we swap out the x for a 1, minus...swap out the x for a 1...squared; 5  2  1 comes out to be 2.1674
So, f(1) comes out to be 2; great.1687
All right, now what we need to figure out is if the limit exists there.1693
If our limit is going to exist, we have two different sides that we are coming from.1698
We are coming from the left side, and we are coming from the right side.1702
We are coming from the less than or equal and the greater than.1706
We have both a left and a right side that we have to make sure match up.1709
Since we have two different possibilities, we have to make sure that the limit as x goes to 1 of f(x) is the same thing as saying1713
(because we have to check that both sides match) that the limit, as x goes to 1 from the negative side,1724
is equal to the limit as x goes to 1 from the positive side.1733
Now, we have to make sure that they are the same; and they should, of course, both be f(x) in here.1741
We have been talking about f(x) this whole time.1747
The limit as x goes to 1 from the negative side: well, actually, we already figured it out.1749
What is the limit as x goes to 1 from the negative side?1753
Well, that is going to be based off of how this works, because it is the left side.1756
x ≤ 1 behaves like 5  2x  x^{2}; so if you are at the less than, you are on the negative side, so it behaves just like 5  2x  x^{2}.1761
So, we already figured out what happens there; that comes out to be 2.1769
Since we are behaving just like the left side, that is just like it is going towards 2.1774
So, we know that this is going to come out to be 2.1779
Really, our only question is, "Does this part here equal the limit as x goes to 1 from the positive side?"1783
We are allowed to determine ax  1; we can't change x, because that is just the variable;1791
but awe are supposed to determine the value of a that will make this whole thing continuous.1796
We know 2, because that is what the limit is as x goes to 1 from the negative side, has to be equal to ax  1.1802
What x are we going to? We are going to 1 from the positive side, so we just use ax  1; we plug in 1 for our x; minus 1.1811
Start working that out: we have 2 = a  1; we add 1 to both sides; we have 3 = a, so 3 must be equal to a.1818
And if 3 is equal to a, then that means that the limit on the right side is equal to the limit on the left side,1826
which means that the limit does exist, and that the limit will come out to be 2.1833
And since the limit comes out to be 2, and we know that f(1) = 2, we now see that, yes, it is, indeed, continuous.1838
It might be a little bit hard to understand what is going on, so it can really help to see this graphically.1845
Let's draw a quick picture, just to cement our understanding.1849
That is how you do it technically; but it is really useful to understand what is going on intuitively, as well.1852
Our 5  2x  x^{2} would graph basically like this.1858
And when we get to x ≤ 1, we jump over to the other one.1867
Now, since it is ax  1, a is taking the position of the slope: mx + b is how we normally graph a line.1871
So, mx  1 means that we are shifted down 1; we are definitely going to have a point there, down one on the yaxis.1881
But our awe are allowed to choose our value of a.1889
So, what we are doing is effectively choosing the slope that we are going to have.1892
So, that means that we get to choose some possibility for our slope.1895
We have any possible rotation of this line; all of the different lines that could go through here with various different slopes are all of the different possibilities.1899
The one that we have to choose to make this thing come out to be continuous1910
is where that line matches up to where we have this handoff, where we have this breakover point.1915
So, we choose the one that matches up; we choose that slope, and it continues out from here.1922
And that way, we end up having that the breakover ends up changing to a new track, but that track starts in the same place.1926
We have chosen the slope so that the line matches up with where the other one finished off; and it goes through; great.1935
All right, that finishes this; we now have an understanding of how to deal with piecewise functions.1941
It is basically a function of if the left side and right side, the onesided limits, match up to each other.1946
And if we are talking about continuity, it is a question of if the limit matches what comes out of it.1952
And sometimes, it will become a question of if the two sides match what comes out of it.1956
All right, that finishes this lesson; we will see you at Educator.com latergoodbye!1960
1 answer
Last reply by: Professor SelhorstJones
Wed Oct 7, 2015 10:51 AM
Post by Andrew Cheesman on September 26, 2015
You truly have a gift... Amazing teacher.
1 answer
Last reply by: Professor SelhorstJones
Sat Nov 9, 2013 4:00 PM
Post by Miguel Cuzquen on November 6, 2013
Where can I get more resolved examples of this topic?
1 answer
Last reply by: Professor SelhorstJones
Sun Apr 14, 2013 6:31 PM
Post by Akilah Miller on April 12, 2013
Thank for the video! It is very well explained and easy to understand.