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Section 3.1 The 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*}
Definition 3.1. 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:

Subsection 3.1.1 One-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:

\begin{equation*} \lim_{x\to 1^-}f(x)=1\qquad\mbox{and} \qquad\lim_{x\to 1^+}f(x)=2\text{.} \end{equation*}

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:

Definition 3.2. 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.

Answer

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

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.

Answer

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

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 \(\ds\lim_{x\to 1^{-}}\frac{|x-1|}{1-x^2}\text{.}\)

Answer

\(\ds\lim_{x\to 1^{+}}\frac{|x-1|}{1-x^2} \approx -0.5\) 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.

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