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$\begingroup$

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.

$\lim_{h\to 0} \dfrac{\frac1{\sqrt{x+h-2}}-\frac1{\sqrt{x-2}}}{h}$

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    If you know differentiation, you can "cheat": Can you see that this is the derivative of $f(x)=1/\sqrt{x-2}$ at $x$?2011-09-13

2 Answers 2

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$\lim_{h\to 0}\dfrac{\dfrac{1}{x+h-2}-\dfrac{1}{x-2}}{h\left(\dfrac{1}{\sqrt{x+h-2}}+\dfrac{1}{\sqrt{x-2}}\right)}$

$ =\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.

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    @Srivatsan Thanks. I edited my answer.2011-09-14
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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*} $

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    @Michael Hardy: You are right. In a similar but a little easier question I [answered](http://math.stackexchange.com/questions/60220/finding-lim-limits-h-to-0-frac-sqrt9h-3h/60227#60227) with the rationalization technique.2011-09-13