2.2: Limits with Tables and Graphs

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Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Given a function $f(x)$ :

The limit $\lim _{x \to a} f(x)$ is the value the function $f(x)$ approaches as $x$ gets closer to $a$ (if it exists).
The left-hand limit $\lim _{x \to a^-} f(x)$ is the value the function $f(x)$ approaches as $x$ gets closer to $a$ for values $x<a$ (if it exists).
The right-hand limit $\lim _{x \to a^+} f(x)$ is the value the function $f(x)$ approaches as $x$ gets closer to $a$ for values $x>a$ (if it exists).

The limit $\lim _{x \to a} f(x)$ exists if and only if $\lim _{x \to a^-} f(x) = \lim _{x \to a^+} f(x)$ .
We write a limit equals $\infty$ (resp. $-\infty$ ) if the values of $f(x)$ get arbitrarily large (resp. arbitrarily large in the negative direction) as $x$ approaches $a$ .

Evaluate $\lim _{x \to 3}\, [(x-1)^2-3]$ .

We will learn a quicker way to computing this limit in the next section, but for now, we have two approaches:

Using a graph. Let $f(x) = (x-1)^2-3$ . Which of the following represents the graph of $y = f(x)$ ?

Based on this graph, it appears that as $x$ gets closer to $3$ , the $y$ -values of $y = f(x)$ get closer to $\answer {1}$ .
Using a table. Using a calculator, fill in the following tables (round to four digits after the decimal point):
$\begin {array}{c|c} x & f(x) \\ \hline 3.1 & \answer {1.41} \\ 3.01 & \answer {1.0401} \\ 3.001 & \answer {1.004} \\ 3.0001 & \answer {1.0004} \\ \end {array}$

$\begin {array}{c|c} x & f(x) \\ \hline 2.9 & \answer {0.61} \\ 2.99 & \answer {0.9601} \\ 2.999 & \answer {0.996} \\ 2.9999 & \answer {0.9996} \\ \end {array}$
Based on these tables, we see that as the $x$ -values get closer to $3$ , the $y$ -values appear to be approaching $\answer {1}$ which doesdoes not match our answer using the graph above.

Calculate $\lim _{x \to 0} \frac {2\sin (x)}{x}$ .

Again, let’s look at both approaches.

Using a graph. The graph of $y = \frac {\sin (x)}{x}$ is shown below.

Notice that this function $f(x) =2 \sin (x)/x$ is undefined at $x=0$ , hence the hole at $x=0$ on the graph. However, as $x$ gets closer to $0$ , we note that the $y$ -values are getting closer to a certain value, and that value is $\answer {2}$ .
This problem illustrates the following: limits care about what is happening around a point, not necessarily what happens at that point.
Using a table. Here is a table showing the values of $f(x)$ for different $x$ -values:
$\begin {array}{c|c} x & f(x) \\ \hline .5 & 1.9177 \\ .1 & 1.99667 \\ .01 & 1.99997 \\ \end {array}$

$\begin {array}{c|c} x & f(x) \\ \hline -.5 & 1.9177 \\ -.1 & 1.99667 \\ -.01 & 1.99997 \\ \end {array}$
Based on these tables, we see that as the $x$ -values get closer to $0$ , the $y$ -values appear to be approaching $\answer {2}$ which doesdoes not match our answer using the graph above.

Consider the following graph of a function $f(x)$ :

Evaluate the following (if they exist). If the quantity does not exist, enter DNE.

$f(-1) = \answer {2}$ .
$\lim _{x \to -1^-} f(x) = \answer {1}$ .
$\lim _{x \to -1^+} f(x) = \answer {2}$ .
Because $\lim _{x \to -1^-} f(x)$ $\lim _{x \to -1^+} f(x)$ , the limit $\lim _{x \to -1} f(x)$ existsdoes not exist .
$f(2) = \answer {1}$ .
$\lim _{x \to 2^-} f(x) = \answer {2}$ .
$\lim _{x \to 2^+} f(x) = \answer {2}$ .
$\lim _{x \to 2} f(x) = \answer {2}$ .
$f(4) = \answer {1}$ .
$\lim _{x \to 4^-} f(x) = \answer {1}$ .
$\lim _{x \to 4^+} f(x) = \answer {1}$ .
$\lim _{x \to 4} f(x) = \answer {1}$ .

Consider the following graph of a function $f(x)$ :

Evaluate the following (if they exist). If the quantity does not exist, enter DNE. If the limit is infinite, write “infinity” (or ”-infinity” for $-\infty$ ).

$\lim _{x \to 2^-} f(x) = \answer {\infty }$ .
$\lim _{x \to 2^+} f(x) = \answer {\infty }$ .
$\lim _{x \to 2} f(x) = \answer {\infty }$ .
$\lim _{x \to 1^+} f(x) = \answer {1}$ .
$\lim _{x \to 1} f(x) = \answer {1}$ .
$\lim _{x \to 0^-} f(x) = \answer {-\infty }$ .
$\lim _{x \to 0^+} f(x) = \answer {\infty }$ .
$\lim _{x \to 0} f(x) = \answer {DNE}$ .
$\lim _{x \to -1} f(x) = \answer {DNE}$ .
$\lim _{x \to -3^+} f(x) = \answer {2}$ .