WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, I thought we would do some more example problems for limits at infinity
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or as x goes to positive or negative infinity.
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Let us jump right on in.
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Evaluate the following limit x² + 6x - 12/ x³ + x² + x + 4, as the limit as x goes to negative infinity.
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Here we have a rational function.
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We know how to deal with rational functions.
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We pretty much just divide the top and bottom, by the highest power of x in the denominator.
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The first thing you do is basically plug in.
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You evaluate the limit to see, before you actually have to manipulate.
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In this case, when we put in negative infinity into here and here, what we are going to end up with is,
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I will say plugging in, we get infinity/ negative infinity.
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x², this term is going to dominate, this term is going to dominate.
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Negative infinity², negative number² is positive.
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Negative number³ is going to be negative.
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We are going to get something like this which does not make sense.
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This is going to be the limit as x goes to negative infinity of x² + 6x - 12/ the greatest power in the denominator which is x³.
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That is our manipulation x³ + x² + x + 4/ x³.
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This gives us the limit as x approaches negative infinity of, here we have 1/ x + 6/ x² – 12/ x³ / 1 + 1/ x + 1/ x² + 4/ x³.
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Now when we take the limit as x goes to negative infinity, this is 0, this is 0, this is 0.
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All of these go to 0 and you are left with 0/1 which is an actual finite number 0.
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Our limit is 0.
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Again, it is always nice to confirm this with a graph.
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This is our function, down here a little table of values.
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Basically, what this part of the table of values does is it confirms the fact that we start negative,
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the function crosses the 0.
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And then actually comes up and gets closer and closer to 0 which is the limit.
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As x gets really big, the function gets close to 0 which is what we just calculated analytically.
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Evaluate the following limit.
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The limit as x goes to infinity of x + 4/ 8x³ + 1.
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A couple of things to notice.
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Again, instead of just launching right in this, you want to stop and ask yourself some questions.
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Here the radical is the denominator.
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It is a rational function.
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We have to think about these things.
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8x³ + 1, it is under the radical sign.
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It, itself has to be greater than 0.
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Let us see what the conditions are here.
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Here the 8x³ + 1 is in the denominator.
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We know that that value cannot be 0.
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It cannot equal 0.
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Since we cannot have the square root of a negative number, 8x³ + 1 has to be greater than 0.
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It could be greater than or equal to 0.
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But then, if it were equal to 0 then you have a 0 in the denominator.
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That is why we have only the relation greater than.
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I will write not greater than or equal to, which normally we could.
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Let us go ahead and work this out first.
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8x³ + 1 is greater than 0.
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We have 8x³ greater than -1.
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We have x³ greater than -1.
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That is greater than -1/8, which implies that x itself has to be greater than -1/2.
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That is our domain.
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We said not that relation, greater than -1/2.
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Here we do not have to worry about going to negative infinity.
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All we have to worry about here is going to positive infinity.
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Because x cannot be, x cannot go to negative infinity.
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We do not have to worry about x going to negative infinity.
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We saved ourselves a little bit of work.
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Our x + 4/ 8x³ + 1, it is a rational function.
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We want to divide it by the greatest power of x in the denominator.
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This is going to be x + 4, x³/ √x³.
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This is going to be x + 4/ x³/2.
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The √x³ is x³/2 / 8x³ + 1/ x³.
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That is going to equal, this is going to be 1/ x ^½.
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This is going to be + 4/ x³/2.
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This is going to be √8 + 1/ x³.
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Now we go ahead and take the limit.
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The limit as x goes to infinity of 1/ x ^½ + 4/ x³/2 / √8 + 1/ x³.
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As x goes to infinity, this goes to 0, this goes to 0, this goes to 0.
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We end up with 0/ √8 which is a finite number.
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Our limit is 0.
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The graph that we get is this.
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We see as x gets really big, the function gets closer and closer and closer to 0.
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There we go, confirmed it graphically.
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Evaluate the following limit.
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The limit of √9x ⁺10 – x³/ x + 5 + 100.
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The best way to handle this is, let us try this.
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Again, my way is not the right way.
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It is just one way, you might come up with, 5 different people might come up with 5 different ways of doing this limit.
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That is totally fine, that is the beauty of this.
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Let us do the following.
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Let us take this 9x ⁺10 – x³.
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As x goes to infinity, here only 9x ⁺10 term is going to dominate.
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I’m just going to deal with that term.
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The same thing for the denominator.
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For the denominator, the x⁵ is the one that I'm going to take.
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Basically, what the limit that I'm going to take is the limit as x approaches infinity of √9x ⁺10/ x⁵.
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Essentially, I just said that this does not matter and this does not matter.
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The limit is going to be essentially the same.
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I just deal with those.
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This is going to equal to the limits as x approaches infinity, √9x ⁺10 = absolute value of 3x⁵.
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Remember, the square root of something is the absolute value of something/ x⁵.
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For x greater than 0, in other words x going to positive infinity.
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The absolute value of 3x⁵/ x⁵, when x is greater than 0, this is just 3x⁵/ x⁵ = 3.
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The limit as x approaches positive infinity of 3 = 3.
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Now for x less than 0, this 3x⁵ absolute value/ x⁵, it is actually going to be -3x⁵/ x⁵ = -3.
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The limit as x goes to negative infinity of -3 = -3.
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Here you are going to end up with two different asymptotes, 3 and -3.
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Now notice the difference between the following.
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We finished the problem but we are going to talk about something.
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Notice the difference between following.
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When I take √x ⁺10, I get the absolute value of x⁵.
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The absolute value of x⁵ is either going to be x⁵ because when x is greater than 0 or it is going to be,
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I actually did it in reverse, that is why I got a little confused here.
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It is going to be –x⁵ and that is when x is less than 0.
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When x is greater than 0, it is going to be x⁵.
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The reason is because this is an odd power.
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x⁵ itself, depending on whether x is positive or negative, this inside is going to be a positive or negative number.
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If what is in here is positive, then it is going to go one way.
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If what is in here is negative, it is going to go the other way.
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However, if I take something like x⁸, this is going to give me,
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we said that the square root of a thing is going to be absolute value of x⁴.
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Here it does not matter.
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If x is positive or negative, it is an even power.
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An even power is always going to be positive number.
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Therefore, this is just going to be x⁴ because it is an even power.
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Be careful of that, you have to watch the powers.
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When you pass from the square root of something, we said the square root of something is the absolute of something.
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But the power itself is going to make a difference on whether you separate or whether you do not.
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Let us take a look at the graph of the function that we just did.
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We said that as x goes to positive infinity, the function approaches 3.
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As x goes to negative infinity, the function approaches -3.
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I did not draw out the horizontal asymptotes here.
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I just want you to see that, but it is essentially what we did.
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Evaluate the following limit, the limit as x approaches positive infinity of 16x² + x.
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You might be tempted to do something like this.
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Let me write this down.
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We are tempted to say as x goes to positive infinity, this term is going to dominate, which is true.
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We are tempted to say that we can treat this as √16x² which is equal to 4x.
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We can just say that 4x - 4x is equal to 0.
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The limit is x approaches this is just 0.
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That is not the case.
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The problem is the x term may contribute to such a point that, what you end up with is infinity – infinity.
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This will go to infinity, this will go to infinity.
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But infinity – infinity, we are not exactly sure about the rates at which this goes to infinity and this goes to negative infinity.
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Because we are not sure about how fast that happens,
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we do not know if it goes to 0 or if it goes to infinity, or if it goes to some other number in between.
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This is an indeterminate form, infinity – infinity.
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We have to handle it differently.
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Let us deal with the function itself, before we actually take the limit.
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Here when we put the infinity in, we get infinity - infinity which does not make sense.
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We have to manipulate it.
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We have 16x² + x under the radical, -4x.
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I’m going to go ahead and rationalize this out.
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I’m going to multiply by its conjugate.
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16x² + x, this is going to be + 4x/ 16x² + x + 4x.
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When I multiply this out, I end up with 16x² + x.
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This is going to be -16x²/ √16x² + x/ +4x.
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Those go away, leaving me with just x/ 16x² + x + 4x.
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We have a rational function, even though we have a square root in the denominator,
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let us go ahead and divide by the largest power in the denominator which is what we always do in the denominator.
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The largest power in the denominator is essentially going to be the √x².
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It is going to be x, but it is going to be √x².
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What we have is the following.
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We are going to have x/x, that is the numerator.
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We are going to have 16x² + x, all under the radical, + 4x all under x, which is going to equal 1/ √16x² + 4x/ x².
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This one, I’m going to treat x as √x².
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I’m going to get 16x² + 4x/ x², + 4x/ x, I’m going to leave this as x.
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This x for these two, because under the radical I’m just going to treat it as √x².
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That ends up equaling 1/ √16, this is x.
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+ 1/ x and this is + 4.
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There we go, now we can take the limit.
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The limit as x approaches positive infinity of 1/ √16 + 1/ x + 4.
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As x goes to infinity, the 1/x goes to 0.
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We are left with 1/ 4 + 4.
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√16 is 4, you will get 1/8.
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Sure enough, that is what it looks like.
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This is our asymptote, this is y = 1/8.
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This is our origin, as x goes to positive infinity, the function itself gets closer and closer to 1/8.
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That is the limit.
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Evaluate the following limit as x goes to infinity, 1 – 5e ⁺x/ 1 + 3e ⁺x.
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When we put x in, we are going to end up with -infinity/ infinity which is in indeterminate form.
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We have to do something with it.
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1 – 5e ⁺x/ 1 + 3e ⁺x, we can do the same thing that we did with rational functions.
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This is going to be the same as, I’m going to divide everything by e ⁺x.
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1 - 5e ⁺x, the top and the bottom, I mean, / 1 + 3e ⁺x/ e ⁺x.
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What I end up with is 1/ e ⁺x - 5/ 1/ e ⁺x + 3.
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I’m going to take the limit of that.
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The limit as x goes to, I’m going to do positive infinity first.
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1/ e ⁺x – 5, put the 1/ e ⁺x + 3.
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As x goes to infinity, e ⁺x goes to infinity that means this thing goes to 0, this thing goes to 0.
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I'm left with -5/3.
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As x goes to positive infinity, my function actually goes to -5/3.
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I have a horizontal asymptote at 5/3, -5/3.
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Now for x going to negative infinity, I have the following.
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The limit as x goes to negative infinity of 1 - 5e ⁺x/ 1 + 3e ⁺x.
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x is a negative number, it is negative infinity.
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e ⁺negative number is 1/ e ⁺positive number.
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This is actually equivalent to the limit as x goes to positive infinity of 1 - 5/ e ⁺x.
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It is e ⁻x is the same as e ⁻x is 1/ e ⁺x.
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Because we are going to negative infinity, x is negative number.
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Because it is a negative number, I can just drop it into the denominator and make it a positive number.
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1 - 5 and then 1 + 3/ e ⁺x.
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As x goes to infinity, this goes to 0, you are left with 1.
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Sure enough, there you go.
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As x goes to negative infinity, we approach y = 1.
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As x goes to positive infinity, our function approaches y = -5/3.
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That is it, just nice manipulation.
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Let us see what we got.
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Now the whole idea of a reachable finite numerical limit is as x gets closer and closer to a certain number or as x goes to infinity,
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but f(x) gets closer and closer to a certain number like we just saw -5/3 or 1.
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This latter number is the limit.
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The question here is how big would x have to be, in order for the function f(x) = e ⁻x/25 + 2
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to be less than a distance of 0.001 away from its limit?
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Closer and closer, closer and closer means we can take it as close as we want.
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In this case, the tolerance that I'm looking for is 0.001 away from its limit.
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The first thing we want to do, what is the limit?
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What is the limit as x goes to infinity of e ⁻x/ 25 + 2.
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Let us just deal with positive infinity here.
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This is the same as e ⁻x/25.
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This is the same as the limit as x goes to positive infinity of 1/ e ⁺x /25 + 2.
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As x goes to infinity, e ⁺x/25 goes to infinity.
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This goes to 0.
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The limit is actually 2.
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The limit of this function as x goes to infinity is equal to 2.
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I probably going to need more room.
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Let me go ahead and go and work in red.
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Now e ⁻x/25 is always greater than 0.
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e ^- x/ 25 + 2 is always going to be greater than 2.
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f(x) which is equal to e ^- x/25 + 2, we said that the limit of this function as x goes to infinity is 2.
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But we said that the function is always greater than 2 which means that
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the function is actually approaching 2 from above.
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It actually looks like this, this is our graph and this is our asymptote at 2.
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The function is doing this.
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The limit is 2, that is this dash line right here.
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We know that the function itself, because this is always greater than 0, the function itself e ⁻x/25 + 2 is always going to be greater than 2.
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It is always going to be above it.
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It is above it, it is getting closer to it from above.
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That is what is happening here physically, getting closer to the 2.
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That is happening from above.
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We want to make this distance, that distance right there.
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We will call it d, we want that distance to be less than 0.001.
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Our question is asking how far out do we have to go?
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What x value passed which x value will this distance?
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This distance between the function and limit be less than 0.001, that is what this is asking.
00:28:00.100 --> 00:28:04.900
Again, we said we will call that d.
00:28:04.900 --> 00:28:14.300
D is equal to the function itself - the limit.
00:28:14.300 --> 00:28:17.900
Here was the limit, here was the function, this is the distance right there.
00:28:17.900 --> 00:28:23.700
We want that distance, that distance is f(x) – l.
00:28:23.700 --> 00:28:28.200
Here is our origin, this is 0,0.
00:28:28.200 --> 00:28:32.400
This number - this number gives me the distance between them.
00:28:32.400 --> 00:28:34.300
It is f(x) – l.
00:28:34.300 --> 00:28:39.400
We know what f(x) is, that is just e ⁻x/25 + 2.
00:28:39.400 --> 00:28:43.900
We know what l is, it is -2.
00:28:43.900 --> 00:28:57.400
These go away, we want this distance which is e ⁻x/25, we want it to be less than 0.001.
00:28:57.400 --> 00:29:04.000
Now we can solve this equation for x.
00:29:04.000 --> 00:29:10.300
I’m going to go ahead and take the natlog of both sides.
00:29:10.300 --> 00:29:24.700
I have -x/25 is less than the natlog of 0.001.
00:29:24.700 --> 00:29:31.000
I’m going to make this a little more clear here, 0.001.
00:29:31.000 --> 00:29:50.200
-x is less than 25 × the natlog of 0.001, that means x is greater than -25 × the natlog of 0.001.
00:29:50.200 --> 00:29:55.100
Whenever I do, the natlog of 0.001 is going to be a negative number.
00:29:55.100 --> 00:30:08.100
Negative × a negative, when I put this in the calculator, I get x has to be greater than 172.069.
00:30:08.100 --> 00:30:10.200
The limit was this number.
00:30:10.200 --> 00:30:13.500
The whole idea of the limit is we want to get closer and closer and closer.
00:30:13.500 --> 00:30:17.100
In this particular case, we specified what we meant by closer and closer.
00:30:17.100 --> 00:30:23.400
I want it closer than 0.001, the function to be less than not far away from the limit.
00:30:23.400 --> 00:30:26.700
I knew that the function was approaching it from above.
00:30:26.700 --> 00:30:32.200
The distance between the function and limit, that is what I want it to be, less than 0.01.
00:30:32.200 --> 00:30:36.400
The distance between the function and limit is the function - the limit.
00:30:36.400 --> 00:30:39.100
This distance, that distance right there.
00:30:39.100 --> 00:30:41.600
I set it and I solve for x.
00:30:41.600 --> 00:30:52.700
As long as x is bigger than 172.69, f(x) - l is going to be less than 0.001.
00:30:52.700 --> 00:31:04.700
In other words, the function is going to be less than 0.001 units away from its limit.
00:31:04.700 --> 00:31:08.300
What if f actually approached it from below?
00:31:08.300 --> 00:31:14.000
What if we have something like this?
00:31:14.000 --> 00:31:17.300
Let us say again, this was our limit.
00:31:17.300 --> 00:31:19.700
This time let us say that the function came from below.
00:31:19.700 --> 00:31:24.600
Now this is f(x) and this is the origin.
00:31:24.600 --> 00:31:28.800
The limit is above the function.
00:31:28.800 --> 00:31:30.700
The distance that we are interested in is this distance.
00:31:30.700 --> 00:31:34.300
The distance between the function and limit.
00:31:34.300 --> 00:31:41.800
Here the distance is going to equal the length - the function.
00:31:41.800 --> 00:31:45.200
The length is a bigger number.
00:31:45.200 --> 00:31:47.000
We want it to be positive.
00:31:47.000 --> 00:31:49.700
It is going to be l – f(x).
00:31:49.700 --> 00:32:07.400
Now we combine f(x) - l and l - f(x) as the absolute value of f(x) – l.
00:32:07.400 --> 00:32:10.500
This distance, and if we are coming from above, this distance,
00:32:10.500 --> 00:32:16.200
they will be the same if we use the absolute value because distance is a positive number.
00:32:16.200 --> 00:32:18.600
You cannot have a negative distance.
00:32:18.600 --> 00:32:24.600
We just combine those two, when we give the definition of a limit by using the absolute value sign.
00:32:24.600 --> 00:32:35.100
It is that absolute value sign that has confounded and intimidated the students for about 150 years now.
00:32:35.100 --> 00:32:54.700
Our formal mathematical definition of the limit.
00:32:54.700 --> 00:32:57.400
We are concerned with the formal mathematical definition of the limit.
00:32:57.400 --> 00:33:04.300
I told you not to believe about that, I do not believe that these kind of definitions,
00:33:04.300 --> 00:33:06.500
these precise definitions do not belong in this level.
00:33:06.500 --> 00:33:08.000
This level is about intuition.
00:33:08.000 --> 00:33:11.900
By intuition, we mean the idea of how close can you get.
00:33:11.900 --> 00:33:15.200
We speak of closer and closer and closer.
00:33:15.200 --> 00:33:17.300
There is a way of describing that symbolically.
00:33:17.300 --> 00:33:21.500
What do we mean by closer and closer, that is what I'm going to describe here.
00:33:21.500 --> 00:33:24.500
Again, I just want you to see it because some of your classes will deal with it,
00:33:24.500 --> 00:33:26.300
some of your classes would not deal with it.
00:33:26.300 --> 00:33:31.400
But I wanted you to see the idea and where it actually came from.
00:33:31.400 --> 00:33:43.200
Our formal mathematical definition of, when we say something like the limit as x approaches infinity of f(x) = l.
00:33:43.200 --> 00:33:46.800
When we say that some function as x goes to infinity,
00:33:46.800 --> 00:33:55.200
that the function actually approaches a finite limit, this symbol, here is what it means mathematically.
00:33:55.200 --> 00:34:12.400
The formal definition is for any choice of a number that we will symbolize with ε,
00:34:12.400 --> 00:34:25.100
which is going to be always greater than 0, there is an x value somewhere on the real line.
00:34:25.100 --> 00:34:35.900
Such that the absolute value of f(x) - the limit is going to be less than this choice of ε,
00:34:35.900 --> 00:34:40.100
whenever x is greater then x₀.
00:34:40.100 --> 00:34:50.000
In the previous example, we found our x₀, that was our 172.
00:34:50.000 --> 00:34:52.700
We found an x₀.
00:34:52.700 --> 00:35:00.200
Our ε in that problem, we chose 0.001 as our ε.
00:35:00.200 --> 00:35:04.100
We wanted to make the difference between the function and the limit less than 0.001.
00:35:04.100 --> 00:35:10.100
We found an x₀ of 172.69.
00:35:10.100 --> 00:35:14.900
Why do I keep writing 6, that is strange.
00:35:14.900 --> 00:35:23.900
Any x value that is bigger than 172.69, we will make the difference between the function and the limit less than 0.001.
00:35:23.900 --> 00:35:31.100
The precise general mathematical definition is, for any choice of the number ε greater than 0,
00:35:31.100 --> 00:35:38.100
there exists an x₀ such that whenever x is bigger than x₀,
00:35:38.100 --> 00:35:42.300
the difference between f(x) and its limit is going to be less than ε.
00:35:42.300 --> 00:35:47.100
You can see why this stuff is confusing and why is it that it actually does not belong at this level.
00:35:47.100 --> 00:35:54.400
Again, for those of you that go on in mathematics and taking analysis course, math majors mostly, this is what you will do.
00:35:54.400 --> 00:35:58.900
You will go back and you actually work with epsilons, deltas, and x₀.
00:35:58.900 --> 00:36:01.900
You will prove why certain things are the way they are.
00:36:01.900 --> 00:36:06.400
At this level, we just want to be able to accept that those proofs had been done.
00:36:06.400 --> 00:36:09.100
We want to be able to use it to solve problems.
00:36:09.100 --> 00:36:11.500
We want to learn how to compute.
00:36:11.500 --> 00:36:13.000
We want to use it as a tool.
00:36:13.000 --> 00:36:14.200
We do not want to justify it.
00:36:14.200 --> 00:36:17.300
Later, you can justify it, as a math major.
00:36:17.300 --> 00:36:19.400
Now we just want to be able to use it.
00:36:19.400 --> 00:36:24.800
This idea of closer and closer and closer to a limit is absolutely fine.
00:36:24.800 --> 00:36:27.800
It is that intuition, if you want to do it.
00:36:27.800 --> 00:36:29.900
Thank you so much for joining us here at www.educator.com.
00:36:29.900 --> 00:36:31.000
We will see you next time, bye.