2
$\begingroup$

I was just wondering if anyone could help me evaluate the following limit, I know the answer I just can't seem to get to it, thanks for any help in advance.

$$\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}.$$

I was calculating the $f'(x)$, where $f(x)=\frac{1}{\sqrt{x}}$ using the definition.

$\endgroup$
2
$\begingroup$

HINT:

Rationalize the numerator to get

$$\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}=-\frac{1}{\sqrt{x+h}\sqrt{x}(\sqrt{x+h}+\sqrt{x})}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

So you have to study the limit when $h\to 0$ of the following quantity : $\frac{\sqrt{x}-\sqrt{x+h}}{h(\sqrt{x}\sqrt{x+h})}$.

Multiplying and dividing by the "conjugate" quantity gives : $\frac{\sqrt{x}-\sqrt{x+h}}{h(\sqrt{x}\sqrt{x+h})} \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} = \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}$ and the limit is straightforward.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy