# 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.

Theorem: [LimitAddition] Suppose that both $$\lim_{x\to x_0}f(x)$$ and $$\lim_{x\to x_0}f(x)$$ exist, and are equal to $$L_f$$ and $$L_g$$, respectively. Then $\lim_{x\to x_0}(f+g)(x)=L_f+L_g.$

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.

Theorem: [LimitProduct] Suppose that both $$\lim_{x\to x_0}f(x)$$ and $$\lim_{x\to x_0}f(x)$$ exist, and are equal to $$L_f$$ and $$L_g$$, respectively. Then $\lim_{x\to x_0}(fg)(x)=L_fL_g.$
We won't prove this for you. The proof is only slightly more technical than the previous one, and is surprisingly similar.

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

Theorem: [LimitReciprocal] Suppose that $$\lim_{x\to x_0}f(x)$$ exists and is equal to $$L_f\ne 0$$. Then $\lim_{x\to x_0} \frac1{f(x)} = \frac1{L_f}.$
We should look at this because it is not quite obvious. We can observe that $\left\vert \frac1{f(x)}-\frac1{L_f}\right\vert = \left\vert \frac{L_f-f(x)}{f(x)L_f}\right\vert.$ We see something that we can make small in the numerator, while the denominator looks more or less like $$L_f^2$$. Thus, the proof is a matter of making the numerator much smaller than $$L_f^2$$. We can do that because $$f(x)$$ approaches $$L_f$$ - we can get as close as we want - but the technical details of showing that can be cumbersome, so we'll skip the actual proof.

Now that we have these theorems, we can prove another dozen easy little results about special cases. For example, if we let $$g$$ be a constant function $$g(x)=c$$, then we see from [ConstantLimit] that $\lim_{x\to x_0} cf(x)=\lim_{x\to x_0} g(x)f(x)=\lim_{x\to x_0} g(x)\lim_{x\to x_0} f(x)=c\lim_{x\to x_0} f(x).$
Theorem: [LimitConstantMultiple] Given any constant $$c$$, if $$\lim_{x\to x_0}f(x)$$ exists and is equal to $$L_f$$, then $\lim_{x\to x_0}cf(x) = cL_f.$

Theorem: [LimitQuotient] Suppose that both $$\lim_{x\to x_0}f(x)$$ and $$\lim_{x\to x_0}f(x)$$ exist, and are equal to $$L_f$$ and $$L_g$$, respectively, with $$L_g\ne0$$. Then $\lim_{x\to x_0}\frac{f(x)}{g(x)} = \frac{L_f}{L_g}.$

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.