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?

$$
\begin{aligned}
\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}
& = \lim _{t\to 0}\left(\frac{\frac{1}{\sqrt{\left(t+8\right)\:+1}}\:-\:\frac{1}{3}}{\left(t+8\right)-8}\right)
\\& = \lim _{t\to 0}\left(\frac{\left(3-\sqrt{t+9}\right)\sqrt{t+9}}{3t^2+27t}\right)
\\& = \lim _{t\to \:0}\left(-\frac{1}{3\left(3+\sqrt{t+9}\right)\sqrt{t+9}}\right)
\\& = \color{red}{-\frac{1}{54}}
\end{aligned}
$$
