# Properties of Limits

Evaluating limits will be much easier when we understand the algebraic properties of limits. Can we add limits? Can we multiply them? We'll just enumerate these properties quickly here.

First, we can add limits.

*Proof:* This result is pretty intuitive -
we believe that you believe it. We'll do a technical
proof just this once, so you can get the flavor of the
arguments.

Since both \(\lim_{x\to x_0}f(x)\) and \(\lim_{x\to x_0}g(x)\) exist by assumption, then for any single choice of \(\epsilon\) we can find two numbers \(\delta_f\) and \(\delta_g\) so that \begin{align*} \vert f(x)-L_f\vert \lt \frac{\epsilon}2\ &\text{whenever}\ \vert x-x_0 \vert\lt \delta_f, \ \text{and} \\ \vert g(x)-L_g\vert \lt \frac{\epsilon}2\ &\text{whenever}\ \vert x-x_0 \vert\lt \delta_g. \end{align*} You will wonder about that \(\epsilon/2\). It is just a notational trick so the the numbers will look pretty in the end. The point is that we can find the appropriate \(\delta\) values no matter how close we require \(f\) and \(g\) to be to the limit, so we are just going to find those \(\delta\) values so that they are twice as close as we might have wanted if we were not adding them together. In fact, we will now set \[\delta = \min\{\delta_f,\delta_g\}.\] In that case, then whenever \(\vert x-x_0\vert\lt \delta\) then we have \begin{align*} \vert (f+g)(x)-(L_f+L_g)\vert &= \vert (f(x)-L_f)+(g(x)-L_g)\vert\\ &\le \vert f(x)-L_f \vert + \vert g(x)-L_g\vert\\ &\lt \frac\epsilon2 +\frac\epsilon2 = \epsilon. \end{align*} Thus, by the definition of the limit, the limit of the sum is equal to the sum of the limits.

A similar result applies for multiplication.

We need one more result that seems slightly less than obvious.

This is again simple: \[ \lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}f(x)\lim_{x\to x_0}\frac{1}{g(x)} =L_f\frac1{L_g}. \] by [LimitProduct]and [LimitReciprocal] in order.

We'll leave other such simple results for you to find.