# Question Video: Finding the Limit of a Difference of Powers Involving Roots Mathematics • 12th Grade

Determine lim_(𝑥 → 4) (√(𝑥 + 12) − 4)/(𝑥 − 4).

04:19

### Video Transcript

Determine the limit as 𝑥 tends to four of the square root of 𝑥 plus 12 minus four all over 𝑥 minus four.

The first thing we should try is direct substitution, just plugging in 𝑥 equals four. If we do this, we get the square root of four plus 12 minus four all over four minus four. Four plus 12 is 16. The square root of 16 is four. And four minus four is zero. So we get the indeterminate form zero over zero. In short, direct substitution doesn’t work here.

What we need to do is to somehow rewrite this fraction in such a way that we can cancel out a common factor of the numerator and denominator. And furthermore, where substituting directly into whatever’s left doesn’t give us zero over zero. So how do we rewrite our fraction? The reason why this fraction is so difficult is because the numerator has this radical in it. It’s hard to find a common factor of the numerator and denominator with this radical here. So what can we do about it?

The key to this question is to multiply both the numerator and denominator by the square root of 𝑥 plus 12 plus four. In other words, it’s the numerator but with the minus sign changed to a plus sign. Of course, multiplying both numerator and denominator by the same thing gives us an equivalent fraction. We can then distribute the terms in the numerator. We distribute the terms in the numerator and we see that the two cross terms cancel. This is not a coincidence. We chose to multiply both numerator and denominator by the square root of 𝑥 plus 12 plus four, in order to make this happen. Here, we were thinking about the difference of squares identity, 𝑎 minus 𝑏 times 𝑎 plus 𝑏 is equal to 𝑎 squared minus 𝑏 squared. As a result, the only term in the numerator involving the square root of 𝑥 plus 12 has it squared. And the square root of 𝑥 plus 12 squared is just 𝑥 plus 12.

So we have succeeded in getting rid of the problematic square root term in the numerator. We have, however, gained a radical in the denominator. But we’ll see soon why this isn’t a problem. We simplify the numerator further to get 𝑥 minus four which cancels with a factor of 𝑥 minus four in the denominator. We’re hoping that this common factor of 𝑥 minus four is the reason that when we substituted in 𝑥 equals four to the fraction, we got the indeterminate form zero over zero. And that having canceled this common factor upon directly substituting the value four in again, we’ll get the value of the limit and not the indeterminate form.

Before we substitute, let’s finish off our algebra. We find that the square root of 𝑥 plus 12 minus four all over 𝑥 minus four is equal to one over the square root of 𝑥 plus 12 plus four. Okay. Let’s clear away our working and get to substituting. Here’s what we showed with our algebra. And this remains true upon taking limits on both sides. We saw before that direct substitution didn’t work on our original fraction but we’re hoping it works on this one.

Replacing 𝑥 by four, we get one over the square root of four plus 12 plus four. Four plus 12 is 16. And the square root of 16 is four. So we get one over eight. So direct substitution did work on the equivalent fraction we found. The main trick in this question was to multiply both numerator and denominator by the square root of 𝑥 plus 12 plus four. Or equivalently, to multiply by the fraction whose numerator and denominator are both the square root of 𝑥 plus 12 plus four. Having done this, we just needed to do some algebra and cancel out the factors of 𝑥 minus four which appeared, before substituting it directly to find the value of the limit.

I call this a trick because this is not something that anyone would necessarily be expected to come up with themselves. This is quite a useful trick. And so, you might want to remember that sometimes it’s useful to rationalize the numerator rather than the denominator.