There are several ways to evaluate a limit:

1. Using a table
2. Based on a graph
3. Using algebra to simplify

We've seen how to do the first two, and in this section, we'll see how we can use algebra to evaluate a limit without needing to build a table or draw a graph (which can be difficult for complicated functions).

Limit Laws

First, let's take a look at some rules that we can use to break limits down into simpler problems. It turns out, though, that we won't need to use these for long, because we'll quickly spot a shortcut. Since we won't actually use these limit laws to solve problems, for the most part, they are mostly included here just to show the foundation; they underlie the work we'll do later, but we won't need to show them after a quick example or two.

These two are pretty straightforward if you look at a graph: the constant function $f(x) = k$ is simply a horizontal line, so no matter what values you plug in or approach, it always returns $k$. Similarly, the identity function $f(x) = x$ returns whatever input you give it, so as you plug in values closer and closer to $c$, you receive back the same values, making the limit just equal to $c$.

Let's see how we can apply these to evaluate limits. Basically, we can use these laws to break complicated expressions down into one of the two basic forms, then apply those definitions over and over.

You may notice something interesting about that last line before the answer, especially if you compare it to the original problem. Before we explore that fully, though, try another one like it.

Notice in both examples that the final line before the answer looks just like it would if we simply evaluated the original function at the point where we're taking the limit.

This actually isn't the full story; this doesn't just apply to polynomials and rational functions, but rather to all continuous functions. We just haven't defined continuity yet, and in fact, we'll use limits to do so, so it would be a bit circular to state that here. Just know that when we encounter continuity a bit later, its definition will be basically that it follows this pattern.

The bottom line, though, is simply this: if you can evaluate the function at $c$, and the function is well-behaved, that's the same as the value of the limit.

As it turns out, this also applies to one-sided limits, which get interesting when we look at piecewise-defined functions. The key there is to think about which definition applies on each side of the point at which we're looking for the limit.

Evaluating Limits Algebraically

What if plugging in the limit point doesn't work? For instance, consider the following limit: $\lim_{x \to 2} \dfrac{x^2-6x+8}{x^2-4}$

Since $x=2$ is not in the domain of this function, the rule we used for the last few examples doesn't apply. In this case, though, we notice that we can simplify the function by factoring and canceling.

$\begin{aligned} \dfrac{x^2-6x+8}{x^2-4} &= \dfrac{(x-4)(x-2)}{(x+2)(x-2)}\ &= \dfrac{(x-4){\color{red}\cancel{(x-2)}}}{(x+2){\color{red}\cancel{(x-2)}}}\ &= \dfrac{x-4}{x+2} \end{aligned}$

Is that really right, though?

Is it true that if $f(x) = \dfrac{x^2-6x+8}{x^2-4}$ we can just write $f(x) = \dfrac{x-4}{x+2}?$

It's almost correct, but there's a small caveat. When we write down a function, we generally assume that the domain is all the real numbers that can be evaluated. So in the first case, we'd notice that $x=2$ and $x=-2$ are both invalid, so the domain doesn't contain either point. In the second case, it only appears that $x=-2$ is not in the domain. So these functions are not exactly equal, because their domains aren't equal. In fact, at every point other than $x=2$, they are equal, but one is implicitly defined there and the other isn't.

The good news

Happily, this doesn't pose a problem when we're looking at limits. Since the limit doesn't see what happens exactly at $x=2$ anyway, we can use the simplified version to evaluate the limit. In essence, when we simplify the function by canceling, we're filling in a hole in the graph.

The punchline

It may seem that we're spending a lot of space on a simple issue, but this is important because most of the limits we'll do in the future will involve some algebraic simplification, so it's important to understand why that works.

If we try to evaluate a limit by plugging in the point in question, and the function returns $\dfrac{0}{0}$ that means there's something that can be cancelled from the top and bottom, and the simplified form will be something we can evaluate: the limits will be the same, so by evaluating the simpler version, we get the answer to the more complicated one we wanted.