<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] Problem:
$$\lim_{x\to-1}\frac{\sqrt{x^2+8}-3}{x+1}$$
Rationalize the numerator by multiplying by its conjugate
$$=\lim_{x\to-1}\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Multiply out the numerator
$$=\lim_{x\to-1}\frac{\sqrt{x^2+8}^2+\boxed{3\sqrt{x^2+8}-3\sqrt{x^2+8}}+-3*3}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Simplify
$$=\lim_{x\to-1}\frac{x^2+8-9}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
$$=\lim_{x\to-1}\frac{x^2-1}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Factor the numerator
$$=\lim_{x\to-1}\frac{\left(x-1\right)\boxed{\left(x+1\right)}}{\boxed{\left(x+1\right)}\left(\sqrt{x^2+8}+3\right)}$$
Cancel
$$=\lim_{x\to-1}\frac{x-1}{\sqrt{x^2+8}+3}$$
Plug in $-1$
$$=\frac{\left(-1\right)-1}{\sqrt{\left(-1\right)^2+8}+3}$$
Simplify
$$=-\frac{1}{3}$$
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