What is a limit?

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We introduce limits.

1 The basic idea

Consider the function

\[ f(x) = \frac {\sin (x)}{x}. \]

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

Use the graph of $f(x) = \frac {\sin (x)}{x}$ 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 $f(x)$ is as close as one wishes to $L$ for all $x$ sufficiently close, but not equal to, $a$.

Use the graph of $f(x) = \frac {\sin (x)}{x}$ above to finish the following statement: “A good guess is that…”

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

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

[Picture]

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): $\lim _{x\to 1} f(x) \begin{prompt}=\answer {-2}\end{prompt}$

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

So just to the right of 2, $f(x)=2$. We cannot find a single number that $f(x)$ approaches as $x$ approaches 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 rounding to three decimal places:

\[ \begin{array}{l|l} x & f(x) \\ \hline 0.1 & \begin{prompt}\answer {0}\end{prompt}\\ 0.01 & \begin{prompt}\answer {0}\end{prompt}\\ 0.001 & \begin{prompt}\answer {0}\end{prompt}\\ 0.0001 & \begin{prompt}\answer {0}\end{prompt}\\ \end{array} \]

We may rush and say that, based on the table above,

\[ \lim _{x\to 0}\sin \left (\frac {\pi }{x}\right )=0. \]

But, recall the definition of the limit: $L$ is the limit if the value of $f(x)$ is as close as one wishes to $L$ for all $x$ sufficiently close, but not equal to, $a$.

From this table we can see that $f(x)(=0)$ is as close as one wishes to $L(=0)$ for some values $x$ that are sufficiently close to $a(=0)$. But this does does not satisfy the definition of the limit, at least, not yet.

But, wait! Fill in another table.

\[ \begin{array}{l|l} x & f(x) \\ \hline 0.3 & \begin{prompt}\answer {-.866}\end{prompt} \\ 0.03 & \begin{prompt}\answer {-.866}\end{prompt} \\ 0.003 & \begin{prompt}\answer {-.866}\end{prompt} \\ 0.0003 & \begin{prompt}\answer {-.866}\end{prompt} \end{array} \]

What do these two tables tell us about

\[ \lim _{x\to 0}\sin \left (\frac {\pi }{x}\right )? \]

$\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right ) = 0$ $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right )=1$ $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right ) = -.866$ $\lim _{x\to 0}\sin \left (\frac {\pi }{x}\right ) = -.433$ The limit does not exist.

The limit does not exist. The first table shows that we can always find a value of $x$ as close as we want to $0$ such that $f(x)=0$. However, the limit is not equal to $0$, since the second table shows that we can also find a value of $x$ as close as we want to $0$ such that $f(x)=-\frac {\sqrt {3}}{2}$. It turns out that for any number $y$, $-1\le y\le 1$, we can find a value of $x$ as close as we want to $0$ such that $f(x)=y$. Check the graph of the function $f$.

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

3 One-sided limits

While we have seen that $\lim _{x\to 2}\lfloor x\rfloor $ does not exist, its graph looks much “nicer” near $a=2$ than does the previous graph near $a=0$. More can be said about the function $\lfloor x\rfloor $ and its behavior near $a=2$.

Intuitively,

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

written

\[ \lim _{x\to a^+} f(x) = L, \]

if the value of $f(x)$ is as close as one wishes to $L$ for all $x>a$ sufficiently close to $a$.

Similarly,

for the function $f$, $L$ is the limit from the left as $x$ approaches $a$,

written

\[ \lim _{x\to a^-} f(x) = L, \]

if the value of $f(x)$ is as close as one wishes to $L$ for all $x<a$ sufficiently close 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 number $x=2$. 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

\[ \lim _{x\to 2^+} f(x)=2. \]

4 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): $\lim _{x\to 1} f(x) \begin{prompt}=\answer {3}\end{prompt}$
(j): $f(0) \begin{prompt}=\answer {-3/2}\end{prompt}$
(k): $\lim _{x\to -3^-} f(x) \begin{prompt}=\answer {6}\end{prompt}$
(l): $\lim _{x\to -2^+} f(x) \begin{prompt}=\answer {2}\end{prompt}$

2025-01-06 19:53:41