AP Calculus AB / Unit 1: Limits and Continuity

1.2 What Is a Limit?

The video for this topic is coming soon. The notes and practice below are ready now.

Notes

The big idea

A limit asks: as x gets closer and closer to a number, what value is f(x) heading toward?

The notation

$\displaystyle \lim_{x \to a} f(x) = L$"The limit of f(x) as x approaches a equals L"

A function with a hole

−112345−11234567

Getting closer to x = 2

From the leftFrom the right
0.1 awayf(1.9) = 3.9f(2.1) = 4.1
0.01 awayf(1.99) = 3.99f(2.01) = 4.01
0.001 awayf(1.999) = 3.999f(2.001) = 4.001

So the limit is 4

$\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2} = 4$even though f(2) is undefined

The algebra way: factor and cancel

  1. $\displaystyle \frac{x^2-4}{x-2}$Plugging in 2 gives 0/0, so we need to simplify
  2. $\displaystyle \frac{(x-2)(x+2)}{x-2}$Factor the top: difference of squares
  3. $\displaystyle x+2$Cancel the (x−2) pieces, since x is never exactly 2
  4. $\displaystyle \lim_{x \to 2} (x+2) = 4$Now plug in 2

Common mistake

$\displaystyle \frac{x^2-4}{x-2} = \frac{0}{0} = 0$
$\displaystyle \frac{0}{0} \;\mathrm{means\ simplify\ first}$

0/0 is a signal, not an answer

Three ways to find a limit

When the sides disagree

−2−11234−11234567(1, 4)

One-sided limits

$\displaystyle \lim_{x \to 1^-} f(x) = 2$from the left
$\displaystyle \lim_{x \to 1^+} f(x) = 4$from the right
$\displaystyle \lim_{x \to 1} f(x) \;\mathrm{does\ not\ exist}$the two sides don't match

Remember for the exam

Practice

From the video
Find $\displaystyle \lim_{x \to 3} \frac{x^2-9}{x-3}$
From the video
If $f(2) = 5$ but $f(x)$ approaches 3 as $x \to 2$, what is $\displaystyle \lim_{x \to 2} f(x)$?
Find $\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2}$.
Find $\displaystyle \lim_{x \to 1} \frac{x^2+2x-3}{x-1}$.
Find $\displaystyle \lim_{x \to -2} \frac{x^2+5x+6}{x+2}$.
Find $\displaystyle \lim_{x \to 0} \frac{x^2+3x}{x}$.
For $f(x) = x+1$ when $x < 1$ and $f(x) = x+3$ when $x \ge 1$, find $\displaystyle \lim_{x \to 1^-} f(x)$.
Using the same $f$, what is $\displaystyle \lim_{x \to 1} f(x)$?
A table shows $f(3.9) = 2.48$, $f(3.99) = 2.498$, $f(4.01) = 2.502$, $f(4.1) = 2.52$. What is the best estimate of $\displaystyle \lim_{x \to 4} f(x)$?

Free response

Write your answer first, then open the scoring guide and check off the points you earned.

Let $f(x) = \frac{x^2-1}{x-1}$ for $x \ne 1$, and $f(1) = 5$.

(a) Find $\lim_{x \to 1} f(x)$. Show your work.

Show the answer and scoring guide

$\displaystyle \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$ for $x \ne 1$.

$\displaystyle \lim_{x \to 1} (x+1) = 2$

Give yourself the points you earned:

(b) Find $\lim_{x \to 1^-} f(x)$ and $\lim_{x \to 1^+} f(x)$.

Show the answer and scoring guide

Both one-sided limits equal $2$, because near $x = 1$ (but not at it) $f(x) = x+1$ from both sides.

Give yourself the points you earned:

(c) Explain why $\lim_{x \to 1} f(x)$ is not equal to $f(1)$.

Show the answer and scoring guide

The limit only depends on values of $f$ near $x = 1$, which head to $2$. The value $f(1) = 5$ is defined separately, so the graph has a hole at $(1, 2)$ and a point at $(1, 5)$.

Give yourself the points you earned: