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Limits - Factoring Method Example 4 Today we’re going to be doing another limits
problem. And, in this one we’re dealing with the function t2+6t+9 divided by t2-9
and we’ve been asked to evaluate the limit as t approaches -3.
So, as with any limit problem, the first thing you want to do is try to plug in the number
you’re approaching, in our case -3, to see if you can get a legitimate answer. And, by
legitimate answer I mean an answer that is not undefined, an answer that is not an indeterminate
form. So, you want to see if you get a good answer because simply plugging in the number
you’re approaching is the easiest way to evaluate a limit and sometimes it works. Sometimes
you get a good answer and you’re done and we don’t want to move on to anything more
complex than that in case we can get away with something easier, just we don’t waste
our time. And, plugging in the number we’re approaching is pretty quick.
So let’s just go ahead and do that quickly and see if we can get an answer and if we
don’t we move on to a more complex form. So, we put in -3 everywhere where we see t.
So -32+6(-3)+9 and then we divide that by -32-9 so when we simplify that we’ll get
9-18+9 divided by 9-9. And if we simplify that, we’ll get 0 over 0 so right there
we got an indeterminate form. It didn’t work so we can’t just evaluate the limit
by plugging in the number we’re approaching; we’ve got to go onto something else.
In this case, our first idea should be to try factoring. Most specifically because our
function is a rational function, it’s a fraction, and, whenever we have a fraction
our first idea, like I said, is going to be to factor because if we can factor either
the numerator and or the denominator and cancel out like factors we might be able to plug
in the number we’re approaching again and get another quick answer. So, we’ll try
factoring, if factoring didn’t work we can move on to another method but of course we’ll
try factoring first. So, sometimes you’ll only be able to factor either the numerator
or the denominator but you should always try to factor both if you can, in our case, looks
like we’re going to be able to factor both. If we factor the numerator we’ll get (t+3)2,
another words t+3 times t+3, and if we factor the denominator we’ll get t+3 times t-3.
So we factored our numerator and denominator. Remember too that when you’re doing a factoring
with limits problem you should always factor both the numerator and denominator as much
as you possibly can. So if you had gotten one of these factors, you know, this t+3 factor
right here, that could be factored further, you should absolutely do that. So, always
factor as far as you can, you know, look at all your factors when you’re done and make
sure that they’re factored as much as they could be factored before you move forward.
So in our case, each of these is factored as much as possible. So now we look to see
if we can cancel out any terms, and in fact we can. We have t+3 in both the numerator
and the denominator so we’ll cancel out one of each. We don’t cancel out this second
t+3 term here because we only have one t+3 term in the denominator so we can only take
one out the denominator and one out of the numerator. So we’re left with t+3 divided
by t-3. And like I said, before you move forward you
should, at this point, plug in the number you’re approaching again to see what kind
of an answer you get. So if we plug in -3 to our answer here we’ll get -3+3 divided
by -3-3 and if we simplify that we’ll get 0 divided by -6 which would just be 0. And,
that’s a good answer. We don’t have an indeterminate form like 0/0 like we did here
and our answer is not undefined because 0 is not in the denominator. The fact that we’ve
got 0 as our limit is a good, clean answer. So, that will be our final answer, we don’t
have to go any further. And, if we’re going to write our final answer
technically, but the way we would write it is this, we would say the limit as t approaches
-3 of t2+6t+9 divided by t2-9 is equal to 0, and all that says is the limit of this
function as t approaches -3 is equal to 0. So, that’s it, that’s our final answer.
Hope that video helped you guys and I’ll see you in the next one.