proof of derivative using definition Use the definition to show that the function $f:[0,+\infty)\to \mathbb R$ such that $f(x)=\sqrt{x}$ for all $x\ge 0$ is differentiable at each $x\in (0,+\infty)$.
My solution is
$x_0= (0,\infty +$)
($\lim_{x\rightarrow x_0}\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0}$)
$\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0}*\frac{\sqrt{x}+\sqrt{x_0}}{\sqrt{x}+\sqrt{x_0}}=
\frac{x-x_0}{(x-x_0)(\sqrt{x}+\sqrt{x_0})}=
\frac{1}{\sqrt{x}+\sqrt{x_0}}$ which as $x$ approaches $x_0= \frac{1}{2\sqrt{x_0}}$
I though I was done but I was told the definition to use was $f(x)-f(x_0)=h(x) (x-x_0)$
Any ideas on where i went wrong?
 A: $\lim_{\Delta x \rightarrow 0} 
\frac{\sqrt{x + \Delta{x}}-\sqrt{x}} {\Delta x} 
\frac{\sqrt{x + \Delta{x}} + \sqrt{x}}{\sqrt{x + \Delta{x}} + \sqrt{x}} =
\lim_{\Delta x \rightarrow 0} 
\frac{{x + \Delta{x}}-{x}} {\Delta x {(\sqrt{x + \Delta{x}}} + \sqrt x )} = \frac{1}{2\sqrt{x}}
$
A: I suspect that you're supposed to use this definition, though I can't quite tell for sure.

Given a set $E\subseteq\Bbb R,$ a point $x_0\in E$ such that $(x_0-c,x_0+c)\subseteq E$ for some $c>0,$ and a function $f:E\to\Bbb R,$ we say that $f$ is differentiable at $x_0$ if there is a number $L$ and a function $h$ defined on $(x_0-c,x_0+c)$ such that $$f(x)-f(x_0)=L\cdot(x-x_0)+h(x)\tag{1}$$ and $$\lim_{x\to x_0}\frac{h(x)}{x-x_0}=0.\tag{2}$$ We say that $L$ is the derivative of $f$ at $x_0$, denoted by $L=f'(x_0).$

Now, you've already figured out that $$f'(x_0)=\frac{1}{2\sqrt{x_0}}$$ for each $x_0\in(0,\infty),$ in your case. Now, we need to show that $(1)$ and $(2)$ hold for some function $h$ when we substitute $L=\frac{1}{2\sqrt{x_0}}.$ Fortunately, since we need $(1)$ to hold after substitution, it's easy to see that we require $$h(x)=f(x)-f(x_0)-L\cdot(x-x_0)=\sqrt{x}-\sqrt{x_0}-\frac{1}{2\sqrt{x_0}}(x-x_0).$$ Then $$\frac{h(x)}{x-x_0}=\frac{\sqrt x-\sqrt{x_0}}{x-x_0}-\frac1{2\sqrt{x_0}}=\frac1{\sqrt x +\sqrt{x_0}}-\frac1{2\sqrt{x_0}},$$ so to show that $(2)$ holds, it suffices to show that $$\lim_{x\to x_0}\frac1{\sqrt x +\sqrt{x_0}}=\frac1{2\sqrt{x_0}}.$$
