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For the function $f(x) = \sqrt{x}, \hspace{2 mm} f: [0, \infty) \rightarrow \mathbb{R}$, I wish to find all points $a$ at which $f(x)$ is differentiable, from the definition $$ f'(x) = \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} $$

Clearly, for our function $f(x) = \sqrt{x}$, this gives us $$ f'(x) = \lim_{x\to a}\frac{\sqrt{x} - \sqrt{a}}{x-a} $$

So the question boils down to showing for what values of $a$ this limit exists. How would this be done? Intuitively, I know that the answer is that the limit exists for all $a \in [0, \infty)$ and is given by $\frac{1}{2 \sqrt{a}}$, but I'm unsure of how to prove this from the limit definition of differentiability.

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    Note that $x-a=(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})$. – 2017-02-09
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    $$ f'(x) \neq \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\mbox{;} \qquad f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\mbox{.} $$ – 2017-02-09

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Note that the derivative is not defined at $x=0$ (see below). By the way, you can write: $$ \lim_{x\to a} \frac{\sqrt{x} - \sqrt{a}}{x-a} = \lim_{x\to a} \frac{1}{\sqrt{x}+\sqrt{a}} $$

From here, it should be easy to discuss when the limit from $x \rightarrow a$ exists and it's finite. Hope this helps.