
evaluating limits
If f(x) = sqrt(x+1) evaluate the following expression:
lim (f(x+h)f(x))/ h
h>0
so my equation looks something like:
(sqrt(x+h+1)sqrt(x+1))/h
how do i simplify the square roots? I'm thinking the h's would cancel out somehow in order for this limit not to be undefined but i don't know how that would happen.

Well you're actually contemplating a derivative via definition of derivative.
Next step: multiply top & bottom by $\displaystyle \sqrt {x + h + 1} + \sqrt {x + 1} .$

$\displaystyle
(\sqrt {x + h + 1}  \sqrt {x + 1} )/ h.
$
the bottom is just h so how would i multiply the top by itself and the bottom by the numerator?

$\displaystyle \lim_{h\to 0} \frac{\sqrt{x+h+1}\sqrt{x+1}}{h}$
$\displaystyle \lim_{h\to 0} \frac{\sqrt{x+h+1}\sqrt{x+1}}{h}\cdot \frac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{ x+1}}$
$\displaystyle \lim_{h\to 0} \frac{(x+h+1)(x+1)}{h(\sqrt{x+h+1}+\sqrt{x+1})}$
$\displaystyle \lim_{h\to 0} \frac{\not h}{\not h(\sqrt{x+h+1}+\sqrt{x+1})}$
$\displaystyle \lim_{h\to 0} \frac{1}{\sqrt{x+h+1}+\sqrt{x+1}}$
Put $\displaystyle h=0$,
$\displaystyle \frac{1}{2\sqrt{x+1}}$