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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (5)

1 answer

Last reply by: Rebecca Dai
Sun Jan 11, 2015 3:42 PM

Post by Bryan Medilien on August 25, 2014

oh my god, this guy is amazing.

1 answer

Last reply by: Angela Patrick
Sun Sep 22, 2013 3:48 PM

Post by Dudley Jackson on August 14, 2013

In example 2    does f(2) equal 17 not 12    ( 2^3 + 9 = 17 )

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Post by Narin gopaul on September 20, 2012

I love you dude you make my life pain free of math that is

Related Articles:

Limits & Continuity

  • Continuity Checklist:
    • Does f(c) exist?
    • Does ?
    • Does ?

Limits & Continuity

f(x) is a continuous function. Given f(12) = −1 and f(20) = 1. Does f(x) = 0 at any point?
  • The key information here is that f(x) is a continuous function. One way a continuous function can be thought of, is that in between any two points on the graph, there are no jumps, skips, or empty spaces. If you were graphing a continuous function with pen and paper, this means that your hand draws a continuous curve without lifting the pen from the paper. It can be inferred that in between two points on a continuous curve, every y value in between exists somewhere on the graph. The same can be said for every x value between those two points. This is called the intermediate value theorem. So yes, f(x) = 0 at some point as it is in between -1 and 1.
Yes, f(x) = 0 at some point as it is in between -1 and 1
Looking at the graph below, find the limx → 0 y
limx → 0 y = 0, This happens to be the graph of y = |x|
Determine if the following function is continuous at x = 0, f(x) = [(|x|)/x]
  • f(−.1) = −1
  • f(−.0001) = −1
  • f(.0001) = 1
  • limx → 0+ does not equal limx → 0, therefore limx → 0 [(|x|)/x] does not exist. NOT CONTINUOUS AT x = 0
NOT CONTINUOUS AT x = 0
Determine if the following function is continuous at x = 0
  • f(x) = x : x ≥ 0 −x : x < 0
  • f(0) = 0
  • limx → 0+ x = 0
  • limx → 0 −x = 0
  • limx → 0 f(x) = 0 = f(0)
  • Yes, the function is continuous at x = 0. This is another way of writing f(x) = |x|
Yes, the function is continuous at x = 0. This is another way of writing f(x) = |x|
Determine if the following function is continuous at x = 2, f(x) = [(x2 − 4)/(x − 2)]
  • f(0) = [(22 − 4)/(2 − 2)] = [0/0] = undefined
  • Not continuous. f(x) must exist for f to be continuous at x
Not continuous. f(x) must exist for f to be continuous at x
Find and identify any discontinuities of f(x) = [1/(2x + 1)]
  • Does the denominator ever equal zero?
  • 2x + 1 = 0
  • 2x = −1
  • 2x ≠ −1
  • There's no exponent you can raise a positive number to to get a negative number. The denominator never equals zero. No discontinuities
There's no exponent you can raise a positive number to to get a negative number. The denominator never equals zero. No discontinuities
Find and identify any discontinuities of f(x) = [(x − 9)/(x2 − 9x)]
  • f(x) = [(x − 9)/(x2 − 9x)]
  • f(x) = [(x − 9)/(x(x − 9))]
  • Discontinuities at x = 0, and x = 9. We can technically factor out the (x − 9), so this is a removable discontinuity. x = 0 is an essential discontinuity, meaning it can't be removed or factored out. In thise case, it is a vertical asymptote at x = 0.
Discontinuities at x = 0, and x = 9. We can technically factor out the (x − 9), so this is a removable discontinuity. x = 0 is an essential discontinuity, meaning it can't be removed or factored out. In thise case, it is a vertical asymptote at x = 0
Find and identify any discontinuities of f(x) = tanx
  • f(x) = tanx = [sinx/cosx]
  • There are going to be vertical asymptotes whenever cosx = 0. So there are asymptotes at x = ...  , −[(π)/2], [(π)/2], [(3π)/2], [(5π)/2], ...
There are going to be vertical asymptotes whenever cosx = 0. So there are asymptotes at x = ...  , −[(π)/2], [(π)/2], [(3π)/2], [(5π)/2], ...
Find and identify any discontinuities of f(x) = [1/(x2 + 14x + 24)]
  • f(x) = [1/(x2 + 14x + 24)]
  • = [1/((x + 12)(x + 2))]
  • Vertical asymptotes at x = −12 and x = −2
Vertical asymptotes at x = −12 and x = −2
Find and identify any discontinuities of f(x) = [1/(ex − 1)]
  • Does the denominator ever equal zero?
  • ex − 1 = 0
  • ex = 1
  • lnex = ln1
  • x lne = 0
  • x = 0
  • Vertical asymptote at x = 0
Vertical asymptote at x = 0

*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

Limits & Continuity

Lecture Slides are screen-captured 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 0:00
  • Definition 0:06
    • 3 Rules: f(x) Is Continuous…
  • Example 1: Finding Continuity 1:06
  • Types of Discontinuity 2:44
    • Jump
    • Point
    • Essential (Asymptote)
    • Removable
  • Example 2: Continuity Examples 4:41
  • Example 3: Continuity Examples 6:13
  • Example 4: Locate & Identify Type of Discontinuities 8:00