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?
 A: 
$$\frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)} = \frac{\color{red}{8-x}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}$$

Careful in the numerator:
$$\frac{1}{x+1}-\frac{1}{9} \ne 8-x$$
but rather:
$$\frac{1}{x+1}-\frac{1}{9}= \frac{9}{9(x+1)}-\frac{x+1}{9(x+1)} = \frac{8-x}{\color{blue}{9(x+1)}}$$
So then after cancelling/simplifying:
$$\frac {\frac{-1}{\color{blue}{9(x+1)}}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \xrightarrow{x \to 8} -\frac{1}{54}$$
A: $$
\begin{aligned}
\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}
& = \lim _{t\to 0}\left(\frac{\frac{1}{\sqrt{\left(t+8\right)\:+1}}\:-\:\frac{1}{3}}{\left(t+8\right)-8}\right)
\\& = \lim _{t\to 0}\left(\frac{\left(3-\sqrt{t+9}\right)\sqrt{t+9}}{3t^2+27t}\right)
\\& = \lim _{t\to \:0}\left(-\frac{1}{3\left(3+\sqrt{t+9}\right)\sqrt{t+9}}\right)
\\& = \color{red}{-\frac{1}{54}}
\end{aligned}
$$
A: 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 $\frac{1}{x+1} - \frac{1}{9}$.
A: $$\lim\limits_{x\rightarrow8}\frac{\frac{1}{\sqrt{x+1}}-\frac{1}{3}}{x-8}=\lim\limits_{x\rightarrow8}\frac{8-x}{3(x-8)\sqrt{x+1}\left(3+\sqrt{x+1}\right)}=-\frac{1}{3\cdot3\cdot6}=-\frac{1}{54}$$
A: Let $\sqrt{x+1}-3=h\implies x=(3+h)^2-1$ 
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}=\lim_{h\to0^+}\dfrac{\dfrac1{h+3}-\dfrac13}{(3+h)^2-9}=\lim_{h\to0^+}\dfrac{3-(h+3)}{3h(6+h)(h+3)}=?$$
Cancel out $h$ safely as $h\ne0$ as $h\to0$
A: By definition, this is $f'(8)$ for $f(x) = (x+1)^{-1/2}.$
