# How to evaluate the limit $\lim_{h \to 0}\frac{h}{\sqrt{h+4}-2}$?

I have tried multiplying the numerator and denominator by $\sqrt{h+4}-2$, but I am struggling to simplify it further. I have solved a few problems which usually end up in cancelling out the common variables, but I am unable to simplify this further.

I haven't learnt about L'Hôpital's rule etc. as I have just started learning calculus.

Hint: as $h \neq 0$, $$\frac{h}{\sqrt{h+4}-2} = \frac{h(\sqrt{h+4}+2)}{h+4-4} = \sqrt{h+4}+2$$

• So the limit is 4? Yes, it is thanks! – 10101010 Oct 30 '17 at 9:24

Hint: Rationalise the denominator by multiplying the numerator and denominator by $\sqrt{h+4} +2$ and cancel the $h's$.

Then you are only left with $\lim_{h \to 0}{\sqrt{h+4}+2}$

The method you'll want to use is multiplying the denominator and numerator both by $\sqrt{h+4} + 2$. Remember to switch the minus to a plus ( or vice versa), otherwise you are just squaring the denominator.

Hint:

The limit is the inverse of

$$\lim_{h \to 0}\frac{\sqrt{4+h}-\sqrt4}{h},$$ which should ring a bell.