Trying to show derivative of $y=x^\frac{1}{2}$ using limit theorem

Question

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.

Answer 1

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

Answer 2

Alternative approach:

Noting that $h$ and $x+h$ need to be positive in order for this to make sense, we have $$ \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{(\sqrt{x+h})^2 - (\sqrt{x})^2} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x} {(\sqrt{x+h}+ \sqrt x)(\sqrt{x+h} - \sqrt{x})} $$