What is a limit?

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We introduce limits.

The basic idea

Consider the function graphed below. While $f(x)$ is undefined at $x=0$ , we can still plot $f(x)$ at other values near $x = 0$ .

Use the graph of $f$ above to answer the following question: What is $f(0)$ ?

$0$ $f(0)$ $1$ $f(0)$ is undefined it is impossible to say

Nevertheless, we can see that as $x$ approaches zero, $f(x)$ approaches one. From this setting we come to our definition of a limit.

Intuitively,

the limit of $f(x)$ as $x$ approaches $a$ is $L$ ,

written $\lim _{x\to a} f(x) = L,$ if the value of $f(x)$ can be made as close as one wishes to $L$ for all $x$ sufficiently close, but not equal, to $a$ .

Use the graph of $f$ above to finish the following statement: “A good guess is that …”

$\lim _{x\to 0}f(x) = 1$ . $\lim _{x\to 1}f(x) = 0$ . $\lim _{x\to 1}f(x) = f(1)$ . $\lim _{x\to 0}f(x) = f(0)$ .

Consider the following graph of $y=f(x)$

Use the graph to evaluate the following. Write DNE if the value does not exist.

(a): $f(-2) \begin {prompt}=\answer {1}\end {prompt}$
(b): $\lim _{x\to -2}f(x)\begin {prompt}=\answer {1}\end {prompt}$
(c): $f(-1) \begin {prompt}=\answer {2}\end {prompt}$
(d): $\lim _{x\to -1}f(x) \begin {prompt}=\answer {2}\end {prompt}$
(e): $f(0) \begin {prompt}=\answer {-2}\end {prompt}$
(f): $\lim _{x\to 0} f(x) \begin {prompt}=\answer {0}\end {prompt}$
(g): $f(1) \begin {prompt}=\answer {DNE}\end {prompt}$
(h): $\lim _{x\to 1} f(x) \begin {prompt}=\answer {-2}\end {prompt}$

Limits might not exist

Limits might not exist. Let’s see how this happens.

Consider the graph of $f(x) = \lfloor x\rfloor$ .

Explain why the limit $\lim _{x\to 2} f(x)$ does not exist.

The function $\lfloor x \rfloor$ is the function that returns the greatest integer less than or equal to $x$ . Recall that $\lim _{x\to 2} \lfloor x \rfloor = L$ if $\lfloor x\rfloor$ can be made arbitrarily close to $L$ by making $x$ sufficiently close, but not equal to, $2$ . So let’s examine $x$ near, but not equal to, $2$ . Now the question is: What is $L$ ?

If this limit exists, then we should be able to look sufficiently close, but not at, $x=2$ , and see that $f$ is approaching some number. Let’s look at a graph:

If we look closer and closer to $x=2$ (on the left of $x=2$ ) we see that $f(x) = 1$ . However, if we look closer and closer to $x=2$ (on the right of $x=2$ ) we see

So just to the right of $x=2$ , $f(x)=2$ . We cannot find a single number that $f(x)$ approaches as $x$ approaches $x=2$ , and so the limit does not exists.

Tables can be used to help guess limits, but one must be careful.

Consider $f(x) = \sin \left (\frac {\pi }{x}\right )$ . Fill in the tables below (to three decimal places): $\begin {array}{l|l} x & f(x) \\ \hline 0.1 & \begin {prompt}\answer {0}\end {prompt}\\ 0.01 & \begin {prompt}\answer {0}\end {prompt}\\ 0.001 & \begin {prompt}\answer {0}\end {prompt}\\ 0.0001 & \begin {prompt}\answer {0}\end {prompt} \end {array} \qquad \text {and}\qquad \begin {array}{l|l} x & f(x) \\ \hline 0.3 & \begin {prompt}\answer {-.866}\end {prompt} \\ 0.03 & \begin {prompt}\answer {-.866}\end {prompt} \\ 0.003 & \begin {prompt}\answer {-.866}\end {prompt} \\ 0.0003 & \begin {prompt}\answer {-.866}\end {prompt} \end {array}$ What do the tables tell us about $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right )?$

$\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right ) = 0$ $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right )=1$ $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right ) = -.866$ $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right ) = -.433$ it is unclear what the tables are telling us about the limit

Neither tables nor graphs can ever tell us for certain what a limit is. However, sometimes they can help “guess” the limit. In this case the graph of $f(x)$ is somewhat more helpful:

We see that $f(x)$ oscillates “wildly” as $x$ approaches $0$ , and hence does not approach any one number.

One-sided limits

While we have seen that $\lim _{x\to 2}\lfloor x\rfloor$ does not exist, more can still be said.

Intuitively,

the limit from the right of $f$ as $x$ approaches $a$ is $L$ ,

written $\lim _{x\to a^+} f(x) = L,$ if the value of $f(x)$ can be made as close as one wishes to $L$ for all $x>a$ sufficiently close, but not equal to, $a$ .

Similarly,

the limit from the left of $f(x)$ as $x$ approaches $a$ is $L$ ,

written $\lim _{x\to a^-} f(x) = L,$ if the value of $f(x)$ can be made as close as one wishes to $L$ for all $x<a$ sufficiently close, but not equal to, $a$ .

Compute: $\lim _{x\to 2^-} f(x)\qquad \text {and}\qquad \lim _{x\to 2^+} f(x)$ by using the graph below

From the graph we can see that as $x$ approaches $2$ from the left, $\lfloor x\rfloor$ remains at $y=1$ up until the exact point that $x=2$ . Hence $\lim _{x\to 2^-} f(x)=1.$ Also from the graph we can see that as $x$ approaches $2$ from the right, $\lfloor x\rfloor$ remains at $y=2$ up to $x=2$ . Hence $\lim _{x\to 2^+} f(x)=2.$

When you put this all together

One-sided limits help us talk about limits.

A limit $\lim _{x \to a} f(x)$ exists if and only if

$\lim _{x \to a^-} f(x)$ exists
$\lim _{x \to a^+} f(x)$ exists
$\lim _{x \to a^-} f(x) = \lim _{x \to a^+} f(x)$

In this case, $\lim _{x \to a} f(x)$ is equal to the common value of the two one sided limits.

Evaluate the expressions by referencing the graph below. Write DNE if the limit does not exist.

(a): $\lim _{x\to 4} f(x) \begin {prompt}=\answer {8}\end {prompt}$
(b): $\lim _{x\to -3} f(x)\begin {prompt}=\answer {6}\end {prompt}$
(c): $\lim _{x\to 0} f(x) \begin {prompt}=\answer {DNE}\end {prompt}$
(d): $\lim _{x\to 0^-} f(x) \begin {prompt}=\answer {-2}\end {prompt}$
(e): $\lim _{x\to 0^+} f(x) \begin {prompt}=\answer {-1}\end {prompt}$
(f): $f(-2) \begin {prompt}=\answer {8}\end {prompt}$
(g): $\lim _{x\to 2^-} f(x) \begin {prompt}=\answer {7}\end {prompt}$
(h): $\lim _{x\to -2^-} f(x) \begin {prompt}=\answer {6}\end {prompt}$
(i): $\lim _{x\to 0} f(x+1) \begin {prompt}=\answer {3}\end {prompt}$
(j): $f(0) \begin {prompt}=\answer {-3/2}\end {prompt}$
(k): $\lim _{x\to 1^-} f(x-4) \begin {prompt}=\answer {6}\end {prompt}$
(l): $\lim _{x\to 0^+} f(x-2) \begin {prompt}=\answer {2}\end {prompt}$