# Compute the Limit: Question 3

• Sep 24th 2011, 09:22 PM
sparky
Compute the Limit: Question 3
Is the following correct?

$\text{Calculate } \lim_{x \rightarrow -1 } \frac{\sqrt{x^2+8}-3}{x+1}$

$= \lim_{x \rightarrow -1 } \frac{\sqrt{x^2+8}-3}{x+1 } \cdot \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}$

$= \lim_{x \rightarrow -1 } \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}$

$= \lim_{x \rightarrow -1 } \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}$

$= \lim_{x \rightarrow -1 } \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}$

$= \lim_{x \rightarrow -1 } \frac{(x-1)}{(\sqrt{x^2+8}+3)}$

$= \frac{(-1-1)}{(\sqrt{1^2+8}+3)}$

$= -\frac{1}{3}$
• Sep 24th 2011, 09:52 PM
TylaYuhh
Re: Compute the Limit: Question 3
I did the same problem on my white board with out looking at your answer and I received the same thing.
• Sep 24th 2011, 10:23 PM
Siron
Re: Compute the Limit: Question 3
• Sep 25th 2011, 03:28 AM
sbhatnagar
Re: Compute the Limit: Question 3
Your answer is entirely correct but you could have done it easier and shorter.

$\lim_{x \to -1 }\frac{\sqrt{x^2+8}-3}{x+1}$

$=\lim_{x \to -1 }\frac{2x}{2\sqrt{x^2+8}}$ (Using L' Hopital's Rule)

$=-\frac{1}{3}$
• Sep 25th 2011, 10:58 PM
Siron
Re: Compute the Limit: Question 3
@sbhatnagar:
You're right, but the problem is that for some reasons l'Hopital is used in cases where it's the only option and avoided in cases where you could solve the limit in a different way.
• Sep 26th 2011, 09:53 AM
HallsofIvy
Re: Compute the Limit: Question 3
And, of course, here we do not know if Sparky has covered L'Hopital's rule or even differentiation yet in his class.