Compute the Limit: Question 3

Is the following correct?

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

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

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

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

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

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

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

$\displaystyle = -\frac{1}{3}$

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.

Re: Compute the Limit: Question 3

Your answer is entirely correct!

Re: Compute the Limit: Question 3

Your answer is entirely correct but you could have done it easier and shorter.

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

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

$\displaystyle =-\frac{1}{3}$

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.

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.