Trying to show derivative of $y=x^\frac{1}{2}$ using limit theorem I am trying to understand why the derivative of $f(x)=x^\frac{1}{2}$ is $\frac{1}{2\sqrt{x}}$ using the limit theorem. I know $f'(x) = \frac{1}{2\sqrt{x}}$, but what I want to understand is how to manipulate the following limit so that it gives this result as h tends to zero:
$$f'(x)=\lim_{h\to 0} \frac{(x+h)^\frac{1}{2}-x^\frac{1}{2}}{h} = \frac{1}{2\sqrt{x}}$$
I have tried writing this as :
$$\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$
But I can't see how to get to the limit of $\frac{1}{2\sqrt{x}}$.
Whilst this is not homework I have actually been set, I would like to understand how to evaluate the limit.

Hint: Simplify
$$
\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}
$$
