Computing calculus limits often involves using an algebraic trick. In this post we’ll look at the example of a ratio of two polynomials. In this case you’ll want to factor the polynomials so that you can eliminate one of the troubling terms. Here’s an example:

lim{x right 2}{{x^2~-~4}/{x~-~2}}

If you just set x = 2, you get 0/0. However this limit is done easily by factoring the numerator. Notice that:

x^2~-~4~=~(x~+~2)(x~-~2)

We can use this to cancel the troublesome term in the denominator. So we get:

lim{x right 2}{{x^2~-~4}/{x~-~2}}~=~lim{x right 2}{{(x~+~2)(x~-~2)}/{x~-~2}} ~=~lim{x right 2}{{(x~+~2)}}

Now we can just set x = 2 and write down the answer:

lim{x right 2}{{(x~+~2)}}~=~4

Sometimes solving calculus limits is a little more tricky. These cases are where, unfortunately, your knowledge of algebra comes in handy. Let’s try:

lim{x right 1}~{{x^4~-~1}/{x^3~-~1}}

The approach to calculating calculus limits of this type is really the same-factor the numerator and denominator when possible. Let’s look at the denominator first, with an odd power it’s a little more troublesome. But we can write it like this:

x^3~-~1~=~(x^2~+~x~+~1)(x~-~1)

Now we can work on the numerator to see how we can get rid of the x – 1 term. Notice that:

x^4~-~1~=~(x^2~+~1)(x^2~-~1)~=~(x^2~+~1)(x~+~1)(x~-~1)

So we have:

lim{x right 1}~{{x^4~-~1}/{x^3~-~1}} ~=~lim{x right 1}~{{(x^2~+~1)(x~+~1)(x~-~1)}/{(x^2~+~x~+~1)(x~-~1)}}

Now we just cancel the x – 1 terms and do the limit:

lim{x right 1}~{{(x^2~+~1)(x~+~1)}/{(x^2~+~x~+~1)}}~=~4/3