Finding limit of $\lim\limits_{h\to 0} \frac1{h}\left(\frac1{\sqrt{x+h-2}}-\frac1{\sqrt{x-2}}\right)$ As expected, if you plug in 0 into the initial equation, the answer is undefined or indeterminate. I tried multiplying the conjugate $\frac1{\sqrt{x+h-2}}+\frac1{\sqrt{x-2}}$ to the numerator and the denominator, but i couldn't simplify this equation enough to avoid the indeterminate value. 
$$\lim_{h\to 0} \dfrac{\frac1{\sqrt{x+h-2}}-\frac1{\sqrt{x-2}}}{h}$$
 A: $$\lim_{h\to 0}\dfrac{\dfrac{1}{x+h-2}-\dfrac{1}{x-2}}{h\left(\dfrac{1}{\sqrt{x+h-2}}+\dfrac{1}{\sqrt{x-2}}\right)}$$
$$ =\lim_{h\to 0} \dfrac{\dfrac{-h}{(x-2)(x+h-2)}}{h\left(\dfrac{1}{\sqrt{x+h-2}}+\dfrac{1}{\sqrt{x-2}}\right)}$$
Now cancel the $h$ and substitute $0$ for h
$$ -\dfrac{1}{(x-2)^2}\cdot \dfrac{\sqrt{x-2}}{2}  = -\dfrac{1}{2(x-2)^{3/2}}$$
I have shown how to "simplify" by multiplying conjugate. Though it would be quicker if you follow Rahul's advise.
A: An alternative evaluation. Let $f(x)=\frac{1}{\sqrt{x-2}}$. Then, by definition of $f'(x)$
$$
\lim_{h\rightarrow 0}\frac{1}{h}\left( \frac{1}{\sqrt{x+h-2}}-\frac{1}{\sqrt{
x-2}}\right) =f^{\prime }(x),
$$
which is
$$
\begin{eqnarray*}
f^{\prime }(x) &=&\left( \frac{1}{\sqrt{x-2}}\right) ^{\prime }=\left( \sqrt{
x-2}^{-1}\right) ^{\prime } \\
&=&-1\times \left( \sqrt{x-2}\right) ^{-2}\times \left( \sqrt{x-2}\right)
^{\prime }=-\frac{1}{x-2}\times \frac{1}{2\sqrt{x-2}} \\
&=&-\frac{1}{2\left( x-2\right) ^{\frac{3}{2}}}.
\end{eqnarray*}
$$
