00:01
So for this limit, we can try direct substitution, since we have a polynomial function divided by another polynomial function.
00:07
So we could try just plugging in negative 1 for t.
00:11
And let's go ahead and do that.
00:12
So we're going to get negative 1 squared plus 6 times negative 1 plus 5.
00:18
And then divided by negative 1 squared plus 3 times negative 1 plus 2.
00:26
Negative 1 squared is 1.
00:28
6 times negative 1 is negative 6.
00:30
Plus 5, and then we have negative 1 squared again, minus 3, and plus 2.
00:36
So our numerator is going to be 1 minus 6, which is negative 5, plus 5 would be 0, and our denominator is 1 minus 3, which is negative 2, plus 2, which is also 0.
00:46
So when we try to actually directly substitute in negative 1 in for t, we see that we get 0 divided by 0.
00:54
So what that tells us is that we're probably going to be able to factor the numerator and the denominator in a way where we'll be able to cancel a term and then we'll be able to evaluate this limit and not get zero for zero...