Here is the problem:

$\displaystyle

\displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}

$

The limit of the denominator is obviously zero, which means that this equation must be converted to find its limit. This can be done my multiplying by the conjugate of the numerator.

$\displaystyle \displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}*\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1}$

$\displaystyle

\displaystyle \lim_{x\to 0 } \frac{1+h-1}{h(\sqrt{1+h}+1)}}

$

The h cancels out, leaving:

$\displaystyle

\displaystyle \lim_{x\to 0 } \frac{1}{\sqrt{1+h}+1}}

$

This is equal to:

$\displaystyle

\frac{1}{\displaystyle \lim_{x\to 0 } {\sqrt{1+h}+\displaystyle \lim_{x\to 0 }1}}

$

Substituting $\displaystyle 0$ for h, $\displaystyle \sqrt{1+h}$ is equal to 1

$\displaystyle

\frac{1}{\displaystyle \lim_{x\to 0 } 1+\displaystyle \lim_{x\to 0 }1}}

$

$\displaystyle \lim_{x\to 0 } 1 = 1$ Thus, the answer is:

$\displaystyle

\displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}=\frac{1}{2}

$