00:01
Okay, given f x is equal to square root of x plus 8 and a is equal to 1.
00:08
Let's find part a, which is m10.
00:12
It's equal to the limit as x approaches a, which is 1 of f of x, which is square root of x plus 8, minus f evaluated at 1, which is square root of 1 plus 8, which gives me square root of 9, which is 3, over x minus a.
00:31
Okay, let's take the conjugate of both sides.
00:42
This is equal to the limit as x approaches 1 of x plus 8 minus 9, which gives me negative 1.
00:51
So x minus 1 over x minus 1 times square roots of x plus 8 plus 3.
00:58
Our x minus 1 is cancel.
01:00
So we get the limit as x approaches 1 of 1 over square roots of x plus 8 plus 3.
01:09
Plugging in our 1, we get 1 over 3...