Question

Evaluate the limit:
\[\lim_{x\to1}\frac{\sqrt{x+8}-3}{x-1}\]
So I know I need to somehow cancel the \(x-1\) or otherwise make the denominator non-zero, since both numerator and denominator is zero now. I'm not sure what operations I can apply though.

Answer 1

id say start by rationalizing num

Answer 2

do u know how to do that?

Answer 3

Yes,
\[(\sqrt{x+8}-3)(\sqrt{x+8}-3)\] gives me \(x-1\). However, the denominator becomes \(-1-3 x+x \sqrt{8+x}\).

Answer 4

emm...be carefull :)\[\frac{\sqrt{x+8}-3}{x-1}\times \frac{\sqrt{x+8}+3}{\sqrt{x+8}+3}\]

Answer 5

Sorry, that was a typo.

Answer 6

so u got\[\frac{\sqrt{x+8}-3}{x-1}\times \frac{\sqrt{x+8}+3}{\sqrt{x+8}+3}=\frac{x-1}{(x-1)(\sqrt{x+8}+3)}\]

Answer 7

Oh, why did I ever combine the denominator and not see the factor... Thanks!

Answer 8

no problem :)