lim(x tends to 3) (square root(x^2+7) -4) divided by (square root(x^2-8) -1)

Question

Pls help and list the steps. Thx

Answer 1

limit ( √(x^2 + 7) - 4)/( √(x^2 - 8) - 1) , as x --> 3

my first step always is to actually sub in the approach value,
sure enough, I get 0/0
That means the "eventually" my expression should factor.
multiply top and bottom by (√(x^2 - 8) + 1)

I peeked at what Wolfram had to say, and it came out with a nice answer
http://www.wolframalpha.com/input/?i=limit+%28+%E2%88%9A%28x%5E2+%2B+7%29+-+4%29%2F%28+%E2%88%9A%28x%5E2+-+8%29+-+1%29++as+x+--%3E+3

let's try L'Hopital's Rule

( √(x^2 + 7) - 4)/( √(x^2 - 8) - 1) , as x --> 3
= lime ( (1/2)(x^2 + 7)^(-1/2) (2x) )/( (1/2)(x^2 - 8)^(-1/2) (2x) ) as x -->3
= lim √(x^2 - 8)/√(x^2 + 7) as x --->3
= √1/√16
= 1/4

Answer 2

thx so much