<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] $$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} = \lim_{x \to -1} \frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)} \\ \\ = \lim_{x \to -1} \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}=\\ \\ \lim_{x \to -1} \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x-1}{\sqrt{x^2+8}+3}=\frac{-2}{6}=-\frac{1}{3}$$
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