<s>[INST] What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the denominator so it always stay in the indeterminate form.
 [/INST] Multiplying
$$\frac{\sqrt{x^2+8}-3}{x+1}$$
by
$$\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}\ \ (=1)$$
gives you
$$\begin{align}\lim_{x\to -1}\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}&=\lim_{x\to -1}\frac{\color{red}{(x+1)}(x-1)}{\color{red}{(x+1)}(\sqrt{x^2+8}+3)}\\&=\lim_{x\to -1}\frac{x-1}{\sqrt{x^2+8}+3}\\&=\frac{-2}{\sqrt 9+3}\\& =-\frac 13.\end{align}$$
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