## Section3.1The Limit

The value a function $f$ approaches as its input $x$ approaches some value is said to be the limit of $f\text{.}$ Limits are essential to the study of calculus and, as we will see, are used in defining continuity, derivatives, and integrals.

Consider the function

\begin{equation*} f(x)=\frac{x^2-1}{x-1}\text{.} \end{equation*}

Notice that $x=1$ does not belong to the domain of $f(x)\text{.}$ Regardless, we would like to know how $f(x)$ behaves close to the point $x=1\text{.}$ We start with a table of values:

\begin{equation*} \begin{array}{l|cccccc} x \amp 0.5 \amp 0.9 \amp 0.99 \amp 1.01 \amp 1.1 \amp 1.5 \\ \hline f(x) \amp 1.5 \amp 1.9 \amp 1.99 \amp 2.01 \amp 2.1 \amp 2.5 \end{array} \end{equation*}

It appears that for values of $x$ close to $1$ we have that $f(x)$ is close to $2\text{.}$ In fact, we can make the values of $f(x)$ as close to $2$ as we like by taking $x$ sufficiently close to $1\text{.}$ We express this by saying the limit of the function $f(x)$ as $x$ approaches $1$ is equal to $2$ and use the notation:

\begin{equation*} \lim_{x\to 1}f(x)=2\text{.} \end{equation*}
###### Definition3.1. Limit (Useable Definition).

In general, we write

\begin{equation*} \lim\limits_{x\to a}f(x)=L\text{,} \end{equation*}

if we can make the values of $f(x)$ arbitrarily close to $L$ by taking $x$ to be sufficiently close to $a$ (on either side of $a$) but not equal to $a\text{.}$

We read the expression $\lim\limits_{x\to a}f(x)=L$ as “the limit of $f(x)$ as $x$ approaches $a$ is equal to $L$”. When evaluating a limit, you are essentially answering the following question: What number does the function approach while $x$ gets closer and closer to $a$ (but not equal to $a$)? The phrase but not equal to $a$ in the definition of a limit means that when finding the limit of $f(x)$ as $x$ approaches $a$ we never actually consider $x=a\text{.}$ In fact, as we just saw in the example above, $a$ may not even belong to the domain of $f\text{.}$ All that matters for limits is what happens to $f$ close to $a\text{,}$ not necessarily what happens to $f$ at $a\text{.}$

Interactive Demonstration. Investigate the limit $\ds\lim_{x\to 1} \dfrac{x^2-1}{x-1}$ using the sliders below:

### Subsection3.1.1One-sided limits

Consider the following piecewise defined function:

Observe from the graph that as $x$ gets closer and closer to $1$ from the left, then $f(x)$ approaches $+1\text{.}$ Similarly, as $x$ gets closer and closer $1$ from the right, then $f(x)$ approaches $+2\text{.}$ We use the following notation to indicate this:

The symbol $x\to 1^-$ means that we only consider values of $x$ sufficiently close to $1$ which are less than $1\text{.}$ Similarly, the symbol $x\to 1^+$ means that we only consider values of $x$ sufficiently close to $1$ which are greater than $1\text{.}$

Interactive Demonstration. Investigate the one sided limits $\ds\lim_{x\to 1^+} f(x)$ and $\ds\lim_{x\to 1^-} f(x)$ using the sliders below:

###### Definition3.2. Left and Right-Hand Limit (Useable Definition).

In general, we write

\begin{equation*} \lim_{x\to a^-}f(x)=L\text{,} \end{equation*}

if we can make the values of $f(x)$ arbitrarily close to $L$ by taking $x$ to be sufficiently close to $a$ and $x$ less than $a\text{.}$ This is called the left-hand limit of $f(x)$ as $x$ approaches $a\text{.}$ Similarly, we write

\begin{equation*} \lim_{x\to a^+}f(x)=L\text{,} \end{equation*}

if we can make the values of $f(x)$ arbitrarily close to $L$ by taking $x$ to be sufficiently close to $a$ and $x$ greater than $a\text{.}$ This is called the right-hand limit of $f(x)$ as $x$ approaches $a\text{.}$

We note the following fact:

Or more concisely:

\begin{equation*} \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L\text{.} \end{equation*}

A consequence of this fact is that if the one-sided limits are different, then the two-sided limit $\ds{\lim_{x\to a}f(x)}$ does not exist, often denoted as DNE.

##### Exercises for Section 3.1.

Use a calculator to estimate $\ds\lim_{x\to 0} \frac{\sin x}{x}\text{,}$ where $x$ is in radians.

$\ds\lim_{x\to 0} \frac{\sin x}{x} \approx 1$

Solution

We can investigate the limit,

\begin{equation*} \lim_{x \to 0} \dfrac{\sin{x}}{x} \end{equation*}

numerically (one fast way would be to use a spreadsheet). Let $f(x) = \dfrac{\sin{x}}{x}\text{.}$

\begin{equation*} \begin{array}{ll} x \amp f(x) \\ \hline 3 \amp 0.047040003 \\ 2 \amp 0.454648713 \\ 1 \amp 0.841470985 \\ 0.5 \amp 0.958851077 \\ 0.25 \amp 0.989615837 \\ 0.025 \amp 0.999895837 \\ 0.0025 \amp 0.999998958 \\ 0.00025 \amp 0.99999999 \\ \end{array} \end{equation*}

Notice that as $x \to 0^{+}\text{,}$ $f(x)$ appears to be approaching $1\text{.}$ To be complete, we need to consider the limit from the negative side,

\begin{equation*} \begin{array}{ll}x \amp f(x) \\ \hline -3\amp 0.047040003 \\ -2 \amp 0.454648713 \\ -1 \amp 0.841470985 \\ -0.5 \amp 0.958851077 \\ -0.25 \amp 0.989615837 \\ -0.025 \amp 0.999895837 \\ -0.0025 \amp 0.999998958 \\ -0.00025 \amp 0.99999999 \\ \end{array} \end{equation*}

As $x \to 0^{-}\text{,}$ $f(x)$ appears again to be approaching $1\text{.}$ Hence, our estimate for $\lim\limits_{x \to 0} \dfrac{\sin{x}}{x}$ is $1\text{.}$

Use a calculator to estimate $\ds\lim_{x\to 0} \frac{\tan(3x)}{\tan(5x)}\text{,}$ where $x$ is in radians.

$\ds\lim_{x\to 0} \frac{\tan(3x)}{\tan(5x)} \approx 0.6$

Solution

First, look at the function behaviour as $x$ approaches $0$ from the right:

\begin{equation*} \begin{array}{c|llll} x \amp 0.1 \amp 0.01 \amp 0.001 \amp 0.0001 \\ \hline \frac{\tan{(3x)}}{\tan{(5x)}} \amp 0.566236 \amp 0.59968 \amp 0.599997 \amp 0.59999997 \end{array} \end{equation*}

So we see that $\lim\limits_{x \to 0^{+}} \dfrac{\tan{(3x)}}{\tan{(5x)}} \approx 0.6\text{.}$ Now look at the left-hand limit:

\begin{equation*} \begin{array}{c|llll} x \amp -0.1 \amp -0.01 \amp -0.001 \amp -0.0001 \\ \hline \frac{\tan{(3x)}}{\tan{(5x)}} \amp 0.566236 \amp 0.59968 \amp 0.599997 \amp 0.59999997 \end{array} \end{equation*}

Therefore $\lim\limits_{x \to 0} \dfrac{\tan{(3x)}}{\tan{(5x)}} \approx 0.6\text{.}$

Use a calculator to estimate $\ds\lim_{x\to 1^{+}}\frac{|x-1|}{1-x^2}$ and $\ds\lim_{x\to 1^{-}}\frac{|x-1|}{1-x^2}\text{.}$

$\ds\lim_{x\to 1^{+}}\frac{|x-1|}{1-x^2} \approx -0.5$ and $\ds\lim_{x\to 1^{-}}\frac{|x-1|}{1-x^2} \approx 0.5$

Solution

We first investigate $\lim\limits_{x \to 1^{+}} \dfrac{|x-1|}{1-x^2}\text{:}$

\begin{equation*} \begin{array}{c|llll} x \amp 1.1 \amp 1.01 \amp 1.001 \amp 1.0001 \\ \hline \frac{|x-1|}{1-x^2}\amp -0.4762 \amp -0.4975 \amp -0.49975 \amp -0.499975 \end{array} \end{equation*}

Therefore we estimate $\lim\limits_{x \to 1^{+}} \dfrac{|x-1|}{1-x^2} \approx -0.5\text{.}$ Similarly, from the left-hand side we have:

\begin{equation*} \begin{array}{c|llll} x \amp 0.9 \amp 0.99 \amp 0.999 \\ \hline \frac{|x-1|}{1-x^2}\amp 0.5263 \amp 0.5025 \amp 0.50025 \end{array} \end{equation*}

and so we estimate $\lim\limits_{x \to 1^{-}} \dfrac{|x-1|}{1-x^2} \approx 0.5\text{.}$

In particular we see that

\begin{equation*} \lim_{x \to 1^{-}} \dfrac{|x-1|}{1-x^2} \neq \lim\limits_{x \to 1^{+}} \dfrac{|x-1|}{1-x^2} \end{equation*}

and so $\lim\limits_{x \to 1} \dfrac{|x-1|}{1-x^2}$ does not exist.