### Video Transcript

Find the limit as π₯ tends to two
of the expression π₯ squared minus four over π₯ squared minus nine multiplied by π₯
minus three over π₯ minus two.

Now, we might be tempted to answer
this question by substituting π₯ equals two into the expression. Letβs see what happens if we go
ahead and do that. Our expression becomes two squared
minus four over two squared minus nine multiplied by two minus three over two minus
two. Now of course, two squared is equal
to four. So the first fraction becomes four
minus four, which is zero, over four minus nine, which is negative five. And thereβs no problems with this
so far.

The second fraction however becomes
negative one over zero. We know that we canβt calculate
negative one divided by zero. Itβs undefined. So weβre going to need to find an
alternative method for solving this problem. Instead, weβre going to look to
simplify our expression by factorising where possible.

Weβll begin by factorising the
expression π₯ squared minus ~~two~~ [four]. This is a special example of the
difference of two squares. We have two square numbers
separated by a subtraction symbol. We therefore find the square root
of each part. The square root of π₯ squared is
π₯. So we add an π₯ at the front of
each bracket. The square root of four is two. So we added two as the numerical
part of each bracket. And with the difference of two
squares, we make one of these plus two and one of these negative two. And weβll see why when we check our
answer.

We can check our answer by
multiplying our brackets back out again. We multiply the first term in each
bracket. π₯ multiplied by π₯ is π₯
squared. We multiply the outer terms. π₯ multiplied by negative two is
negative two π₯. Then, the inner terms, two
multiplied by π₯ is two π₯. And the last terms, two multiplied
by negative two is negative four. And we can see now that negative
two π₯ plus two π₯ is zero. So the expansion of π₯ plus two by
π₯ minus two does indeed give us π₯ squared minus four as required.

Weβll repeat this process for π₯
squared minus nine. Once again, we have an π₯ at the
front of each bracket. And the square root of nine is
three. So this factorizes to π₯ plus three
multiplied by π₯ minus three. And so now, weβre finding the limit
as π₯ tends to two of π₯ plus two multiplied by π₯ minus two over π₯ plus three
multiplied by π₯ minus three multiplied by π₯ minus three over π₯ minus two.

And now, we can simplify. We divide through by π₯ minus
two. We then divide through by π₯ minus
three. And we can see that weβre now
multiplying π₯ plus two over π₯ plus three by one. And we have the limit as π₯ tends
to two of π₯ plus two over π₯ plus three. And now, we can evaluate this. Letβs substitute two into our
expression. We get two plus two over two plus
three, which is four over five.

And we see that the limit as π₯
tends to two of π₯ squared minus four over π₯ squared minus nine multiplied by π₯
minus three over π₯ minus two is four-fifths.