B.3 Factor to Find a Limit

Let's see how we can frequently factor to find a limit. This is the tactic you will use most often to compute limits, so practice using our problems with complete solutions.

We'll begin by continuing with the example we introduced at the bottom of the previous screen, and see how we can use the simple algebraic tactic of factoring to find the limit we're after.

Example 1: 𝑓(𝑎) =𝑥2−4𝑥−2.

Find lim𝑥→2𝑥2−4𝑥−2.

Solution.

Step 1. Try Substitution.
We always try Substitution first, as we did at the bottom of the preceding screen: lim𝑥→2𝑥2−4𝑥−2?=4−42−2=00⟹indeterminate result where "indeterminate result" means "we don't know what the value actually is, and hence have more work to do."

Step 2. Factor and simplify.
Let's factor the numerator of our function: lim𝑥→2𝑥2−4𝑥−2=lim𝑥→2(𝑥+2)(𝑥−2)𝑥−2 Ah, now we can cancel the problematic term: lim𝑥→2𝑥2−4𝑥−2=lim𝑥→2(𝑥+2)(𝑥−2)𝑥−2=lim𝑥→2(𝑥+2)

Step 3. Now easy Substitution to finish: lim𝑥→2𝑥2−4𝑥−2=2+2=4✓

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Why does this approach work?

Notice that the function 𝑥2−4𝑥−2 simplified to 𝑥 +2 when we factored it. The only difference between the two functions is that 𝑥2−4𝑥−2 is not defined at 𝑥 =2, since the denominator is zero there, whereas 𝑥 +2 is defined everywhere. To illustrate that 𝑥2−4𝑥−2 is undefined at 𝑥 =2, we show a hole in its graph at that point in the left-hand figure; at every other point, the graph is exactly the same as the the graph of 𝑓(𝑥) =𝑥 +2, shown in the right-hand figure.

By comparing the two graphs, you can see why the limits are the same: whether or not the function is defined at 𝑥 =2, as we approach that value we get closer and closer to the height 𝑦 =4. The whole concept of the limit was created for just this situation, so we can imagine getting as close as we'd like to 𝑥 =2 from the left or from the right without ever fully reaching that point. The process of factoring simply lets us rewrite the function in a form to which we can apply Substitution and conclude that

lim𝑥→2𝑥2−4𝑥−2=lim𝑥→2(𝑥+2)=4

[If you'd like to revisit the interactive Desmos graph for this limit, it's the very first one we looked at in our Limits Introduction.]

As you will see in many, many problems to come, we will use this basic technique often: through algebraic manipulation, we are able to rewrite a function that isn't defined at the point of interest and "turn it into" a different form that is defined at the point of interest. Simple substitution then works to know the value of the function at that point, which in turn is the limit of the function at that point.

The rewritten function is not the same as the initial function

There is an important subtlety here: the two functions in the graphs above are not the same because they have different domains: the original function, 𝑥2−4𝑥−2, is not defined at 𝑥 =2, whereas our rewritten function is. At every point other than 𝑥 =2, they are identical, but because they behave differently at this one point they are not identical functions. We can make them identical, however, by simply excluding this one point: 𝑥2−4𝑥−2=𝑥+2for 𝑥≠2 And because we explicitly don't care about what's happening at 𝑥 =2, and instead only what happens close to 𝑥 =2, we have lim𝑥→2𝑥2−4𝑥−2=lim𝑥→2(𝑥+2)=4

More generally, we have the incredibly helpful rule:

The rule merely encapsulates the reasoning we used above, and will use time and again.

This is almost the only time we'll discuss this subtlety. But please keep in mind that the algebraic manipulations you'll do in almost every problem allow us to find the limit of the original function only because of the way the limit is defined to focus on the behavior of the function close to, rather than at, the point of interest.

You saw in Example 1 above the three simple steps for the tactic of factoring to find a limit:

Practice the tactic of factoring to find the limit in the next few problems. These are straightforward once you learn to recognize what to do.

Practice Problems: Factor to Find a Limit

Every Calculus exam that we've ever seen that includes limits as a topic has at least one problem that requires this tactic. Practice accordingly to be prepared.

Time to practice, especially because you can be sure that you'll see a problem like this on your next exam.

The Upshot

1. If when you try Substitution you obtain 00, then if possible factor the numerator and/or denominator, and reduce the fraction. Almost always, Substitution will then work.

On the next screen we'll explain another tactic you'll need often to compute a limit, using conjugates.

Does factoring as a tactic help you solve limit problems? Do you have additional tips to share? Let our learning community know by posting on the Forum!