00:01
We have the limit as t approaches 2 of t cubed plus 3 t squared minus 12 t plus 4 over t cubed minus 4 t.
00:10
First option would be to do direct substitution.
00:14
I'll fill 2 in for t.
00:16
That will give me 8 plus 12 minus 24 plus 4 on top and 8 minus 8 on bottom.
00:27
That gives me 0 on top, 0 on bottom, which is an indeterminate.
00:34
This doesn't divide out evenly, so what i'm going to do is factor out the denominator first.
00:43
I could pull out a t as a common factor, which would give me t times t squared minus 4.
00:53
And t squared minus 4 is actually a difference of squares.
00:56
So let's make that t plus 2 times t minus 2.
01:02
T plus 2 times t minus 2.
01:14
And we still have that on top.
01:22
Now currently it's not readily apparent that we have anything on top that cancels with the bottom.
01:28
However, i don't want to end up with an indeterminate of 0 over 0.
01:33
The reason the t minus 2 is 0.
01:42
So what i'm going to do is see if that t minus 2 divides out into the top evenly.
01:53
So let's go t cubed plus 3 t squared minus 12 t plus 4 divided by t minus 2.
02:04
T times t squared is t cubed.
02:07
Negative 2 times t squared...
