How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it. 
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$

\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?

We have $$\begin{align}\frac{\frac{1}{x+1} - \frac{1}{9}}{(x-8)((x+1)^{-1/2} + 1/3)} &= \frac{\frac{8-x}{9(x+1)}}{(x-8)((x+1)^{-1/2} + 1/3)} \\&= -\frac{1}{9(x+1)((x+1)^{-1/2} + 1/3))} \\ &\longrightarrow-\frac{1}{9(9)(2/3)} = \color{blue}{-\frac{1}{54}}.\end{align}$$
In particular you forgot to take into account the $9(x+1)$ term you get when simplifying $\frac{1}{x+1} - \frac{1}{9}$.
