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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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You know that $\frac a b - \frac c d = \frac{ad - bc}{bd}$, right? – Rahul Sep 13 '11 at 7:39
If you know differentiation, you can "cheat": Can you see that this is the derivative of $f(x)=1/\sqrt{x-2}$ at $x$? – Fredrik Meyer Sep 13 '11 at 7:46

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

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

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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kuch nahi, Nice solution, but a slight nitpick in terminology. You do not substitute the limit; the limit is the final answer. What you substitute is $0$ in place of $h$. :-) – Srivatsan Sep 13 '11 at 19:40
@Srivatsan Thanks. I edited my answer. – kuch nahi Sep 14 '11 at 6:57

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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This works, but usually limits like this one are just exercises in using the definition of the derivative, which is then used to justify things like the product rule, etc., and those are then used for finding derivatives the way you did. Lots of students fail to appreciate that there can be some circular reasoning in using L'Hopital's rule to prove that $(d/dx)(x^n) = nx{n-1}$, and the like. So I dislike doing too much of this kind of thing. – Michael Hardy Sep 13 '11 at 16:18
@Michael Hardy: You are right. In a similar but a little easier question I answered with the rationalization technique. – Américo Tavares Sep 13 '11 at 16:37

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