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.

Multiply the function by $$\frac{\sqrt{x^2 + 8} +3}{\sqrt{x^2 + 8}+3}$$
You'll have a difference of squares in the numerator of the form $(a -b)(a+ b)$ which, of course, is $a^2 - b^2$. You should have gotten, in the numerator: $$(\sqrt{x^2 + 8})^2 - (3^2) = x^2 + 8 - 9 = x^2 - 1 = (x+1)(x-1)$$
Simplifying, you'll get  $$\lim_{x\to -1} \frac {\overbrace{x^2 -1}^{(x + 1)(x-1)}}{(x+1)(\sqrt{x^2 + 8} +3)} = \lim_{x\to -1} \frac {x-1}{\sqrt{x^2 + 8} +3} = \frac{-2}{6} = -\frac 13$$
