# Limit Laws

When working with limits, it helps to understand how they behave under different circumstances. Limit laws give us a simple framework to be able to understand what we can do to simplify limits. You can think of the limit laws like a toolbox for solving limit problems. For each problem, you may need to apply one or more limit laws in order to be able to solve it successfully.

**Limit Laws: **Suppose that c is a constant, f(x) and g(x) are functions, and the limits defined below exist. If this is true, then the following limit laws hold:

- $lim_{x \to a} (f(x)+g(x)) = lim_{x \to a} f(x) + lim_{x \to a} g(x)$
- $lim_{x \to a} (f(x)-g(x)) = lim_{x \to a} f(x) - lim_{x \to a} g(x)$
- $lim_{x \to a} (cf(x)) = c*lim_{x \to a} f(x)$
- $lim_{x \to a} (f(x)*g(x)) = lim_{x \to a} f(x) * lim_{x \to a} g(x)$
- $lim_{x \to a} \frac{f(x)}{g(x)} = \frac{lim_{x \to a} f(x)}{lim_{x \to a} g(x)}$
- $lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{lim_{x \to a} f(x)}$, where n is positive

These laws are mostly intuitive, they allow us to separate functions from limits, such that we can apply the limit separately, and then perform the operation on the results. There are a few other laws that will be helpful for evaluating limits.

**Power Law: **$lim_{x \to a} (f(x))^n = (lim_{x \to a} f(x))^n$

This definition follows logically from the 4^{th} limit law. If we let f(x) and g(x) be equal to each other, then we would have a square, or power 2. From this, we can see that a squared function would be the same as the limit result squared. Of course, any other exponent is also repeated multiplication, so the same logic applies for any power, hence the power law.

**Constant Law: **$lim_{x \to a} c = c$

The limit of a constant will always be just that constant. Of course, a constant will always have the same value, meaning no matter where x is, c never changes.

**Special Limits: ** The following functions have a defined value for limits:

- $lim_{x \to a} x = a$
- $lim_{x \to a} x^n = a^n$
- $lim_{x \to a} \sqrt[n]{x} = \sqrt[n]{a}$

These functions are special because they are functions that are defined on their whole domain. Since the functions are defined at the point we are evaluating the limit at, the limit is equal to that value, since both sides will always approach the defined value. More broadly defined, the direct substitution property covers all functions as follows.

**Direct Substitution Property: **If f is a polynomial or a rational function, and a is in the domain of f, then $lim_{x \to a} f(x) = f(a)$

The direct substitution property is very helpful, as many limits may contain polynomial and rational functions, which can be easily solved if our limit value is in the domain.

One final property that is very valuable has to deal with equivalent functions, and simplification.

**Defintion: **If $f(x) = g(x)$ when $x \neq a$, then $lim_{x \to a}f(x) = lim_{x \to a} g(x)$, provided the limit exists.

Let's take a look at an example to help understand how these laws and properties can be applied.

**Example: **Find $lim_{x \to 1} \frac{x^2-1}{x-1}$

First, let's try to use the direct substitution property. At x = 1, we get: $\frac{1^2-1}{1-1} = \frac{0}{0}$, which is undefined.

So, let's try to apply some laws to help solve this problem. First, let's factor the upper function to see if we can find some way to remove the undefined behavior. We can simplify $x^2 - 1$ into $(x-1)(x+1)$, which gives us the function $lim_{x \to 1} \frac{(x-1)(x+1)}{(x-1)}$

This in turn allows us to simplify to get a function that is equal to the original function, with the exception of the point x=1. From here, we can apply direct substitution on the result:

$lim_{x \to 1} \frac{(x-1)(x+1)}{(x-1)}$

$= lim_{x \to 1} \frac{\textbf{(x-1)}(x+1)}{\textbf{(x-1)}}$

$= lim_{x \to 1} (x+1)$

$= 1+1 = 2$

It is helpful to go back and look at which properties allowed us to complete this limit problem. Since (x+1) is equal to $\frac{(x-1)(x+1)}{(x-1)}$ at all points except for 1, we can apply the defintion above to continue with the limit. From here, (x+1) is a polynomial, so we can use direct substitution to solve the limit.