The Definition - Examples

Example: [QuadraticLimit] Use the definition to show that $\lim_{x\to 1} x^2-1 = 0.$

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.

1. Choose $$x>x_0$$, and look at $$\vert f(x)-L\vert$$ for that case. This gets rid of one set of absolute values.
2. 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.
3. Repeat the previous steps for the case $$x_0-\delta\lt x\lt x_0.$$
4. Choose the smaller of the $$\delta$$ choices.

Example: [ConstantLimit] If $$g$$ is a constant function $$g(x)=c$$, then $\lim_{x\to x_0} g(x) = c$

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.

Example: [DifferenceQuotientLimit] Use the definition to evaluate $\lim_{x\to1}\frac{x^2-1}{x-1}$

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 $$\lim_{x\to1} \frac{x^2-1}{x-1} = 2$$

Example: [SineLimit] Use the definition to show that $\lim_{x\to 0}\frac{\sin x}{x}=1$

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.