# The Definition - Examples

We must show that no matter what choice we make for \(\epsilon\), we can find a value for \(\delta\) so that whenever \(\vert x-1\vert<\delta\) it follows that \(\vert (x^2-1)-0\vert<\epsilon\). Now, when \(x>1\) and \(\vert x-1\vert<\delta\), then \(1\lt x\lt 1+\delta\) and \begin{align*} \vert (x^2-1)-0\vert &= x^2-1\\ &< (1+\delta)^2-1\\ &= 1+2\delta+\delta^2-1\\ &= 2\delta+\delta^2. \end{align*} Thus, if we choose \(\delta=\frac\epsilon3\), then so long as \(\epsilon\lt 1\) then \(2\delta+\delta^2<\frac23\epsilon+\frac13\epsilon=\epsilon.\) On the other hand, if \(x\lt 1\) and \(\vert x-1 \vert<\delta\), then \(1-\delta\lt x\lt 1\) and \begin{align*} \vert (x^2-1)-0\vert &= 1-x^2\\ &\lt 1-(1-\delta)^2\\ &= 1-(1-2\delta+\delta^2)\\ &= 2\delta-\delta^2. \end{align*} This time if we choose \(\delta=\frac\epsilon3\) then \(2\delta-\delta^2=\frac23\epsilon-\frac{\epsilon^2}9\lt \epsilon.\) We conclude that whenever we choose \(\delta=\frac\epsilon3\) then \(\vert (x^2-1)-0\vert<\epsilon,\) i.e. we can make \(x^2-1\) as close as we want to zero just by choosing \(x\) close to 1.

It is worth observing the steps here, just in case you have to do this yourself some time.

- Choose \(x>x_0\), and look at \(\vert f(x)-L\vert\) for that case. This gets rid of one set of absolute values.
- Use \(x_0\lt x\lt x_0+\delta\) to write \(\vert f(x)-L\vert\) so as to choose a \(\delta\) small enough that \(\vert f(x)-L\vert<\epsilon\) always. Sometimes this means solving for \(\delta\) in terms of \(\epsilon\); more often it means finding an inequality relating \(\delta\) and \(\epsilon\) and then choosing \(\delta\) small enough.
- Repeat the previous steps for the case \(x_0-\delta\lt x\lt x_0.\)
- Choose the smaller of the \(\delta\) choices.

This is exceptionally intuitive, but we will use the result later, so let's see why it seems so obvious. For any positive \(\epsilon\), we can see that \[\vert g(x)-c\vert = 0 < \epsilon,\] regardless of how we choose \(\delta\). That's all we need - choose \(\delta\) any way you want, and the definition is satisfied.

This is a very common situation in Calculus. We have
an expression of the form \( (f(x)-f(x_0))/(x-x_0)\).
This is always the case for computing a derivative, which
will be important in all that follows. Here we evidently
have \(x_0=1\). Computing the limit is easy, but it is
important to remember that whenever we compute limits
we are dealing with points near \(x_0\), but we never
let \(x\) *be* \(x_0\). Thus we can assume that
\(x-1\ne 0\), hence
\begin{align*}
\frac{x^2-1}{x-1} &= \frac{(x-1)(x+1)}{x-1}\\
&= x+1.
\end{align*}
We can conclude that
\(\lim_{x\to1} \frac{x^2-1}{x-1} = \lim_{x\to1} (x+1).\)
We can guess that the limit we want is 2.
When \(1\lt x\lt 1+\delta\) then
\(\vert (x+1)-2\vert=x-1\lt 1+\delta-1\).
If we choose \(\delta=\epsilon\) then
\(\vert (x+1)-2\vert < \epsilon.\)
In the case that
\(1-\delta\lt x\lt 1\) we have
\begin{align*}
\vert (x+1)-2\vert &=\vert x-1\vert\\
&= 1-x\\
&\lt 1-(1-\delta)=\delta.
\end{align*}
We can again choose \(\delta=\epsilon\).
Thus, we conclude that whenever
\(\vert x-1\vert<\epsilon\) then
\(\vert (x+1)-2\vert < \epsilon\), i.e.
that \(\lim_{x\to1}(x+1)=2.\)
Since
\(\lim_{x\to1} \frac{x^2-1}{x-1} = \lim_{x\to1} (x+1)\)
this means that
\(\lim_{x\to1} \frac{x^2-1}{x-1} = 2\)

It turns out that this is a very important limit in Calculus, but some of the details are quite technical, so we have relegated them to an appendix.

Now our aim is to squeeze \(\sin x/x\) between \(\cos x\) and 1. The idea is that as \(x\) approaches zero, then by [CosineZeroLimit] \(\cos x\) approaches 1, so if \(\sin x/x\) stays between those two quantities for all \(x\) then it must also approach 1. It is easy to see that \[\frac{\sin x}{x}\lt \frac{x}x = 1,\] because [SineCompare] shows that \(\sin x\lt x\). On the other hand, since we are dealing with a limit, we are not interested in the case when \(x=0\) so we can assume that \(x\ne 0\), which means that \[\cos x = \frac{\sin x}{\tan x}\lt \frac{\sin x}x,\] because [TanCompare] shows that \(x\lt\tan x\).

Now we can say simply that \[\lim_{x\to0}\cos x \le\lim_{x\to0}\frac{\sin x}x\le \lim_{x\to0}1.\] Since the first and last limits are both 1 and the middle one is squeezed between those, it must also be 1.