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.
 A: 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  $$\lim_{x\to -1} \frac {\overbrace{x^2 -1}^{(x + 1)(x-1)}}{(x+1)(\sqrt{x^2 + 8} +3)} = \lim_{x\to -1} \frac {x-1}{\sqrt{x^2 + 8} +3} = \frac{-2}{6} = -\frac 13$$
A: Use L'hopital's rule (Valid since the function is of the form $\frac{0}{0}$). The derivative of the top has limt equal to $-\frac{1}{3}$ and the derivative of the denominator is the constant $1$.
A: One way to solve the problem is to consider the function
$$f(x)=\sqrt{x^2+8}$$
so the desired limit is 
$$\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=f'(-1)=\frac{x}{\sqrt{x^2+8}}\Bigg|_{x=-1}=-\frac13$$
A: Multiplying
$$\frac{\sqrt{x^2+8}-3}{x+1}$$
by
$$\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}\ \ (=1)$$
gives you
$$\begin{align}\lim_{x\to -1}\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}&=\lim_{x\to -1}\frac{\color{red}{(x+1)}(x-1)}{\color{red}{(x+1)}(\sqrt{x^2+8}+3)}\\&=\lim_{x\to -1}\frac{x-1}{\sqrt{x^2+8}+3}\\&=\frac{-2}{\sqrt 9+3}\\& =-\frac 13.\end{align}$$
A: we have
$\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}$
simplifying this we get
$\lim_{x \to -1}\frac{x-1}{\sqrt{x^2+8}+3}$
Sonnhard.
A: $$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} = \lim_{x \to -1} \frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)} \\ \\ = \lim_{x \to -1} \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}=\\ \\ \lim_{x \to -1} \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x-1}{\sqrt{x^2+8}+3}=\frac{-2}{6}=-\frac{1}{3}$$
A: 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
$$=-\frac{1}{3}$$
