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.
