Dgdrxrt

Use the definition to show that the function $f:[0,+infty)to mathbb R$ such that $f(x)=sqrtx$ for all $xge 0$ is differentiable at each $xin (0,+infty)$.
My solution is

$x_0= (0,infty +$)

($lim_xrightarrow x_0fracsqrtx-sqrtx_0x-x_0$)

$fracsqrtx-sqrtx_0x-x_0*fracsqrtx+sqrtx_0sqrtx+sqrtx_0=
fracx-x_0(x-x_0)(sqrtx+sqrtx_0)=
frac1sqrtx+sqrtx_0$ which as $x$ approaches $x_0= frac12sqrtx_0$

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?

$lim_Delta x rightarrow 0
fracsqrtx + Deltax-sqrtx Delta x
fracsqrtx + Deltax + sqrtxsqrtx + Deltax + sqrtx =
lim_Delta x rightarrow 0
fracx + Deltax-x Delta x (sqrtx + Deltax + sqrt x ) = frac12sqrtx
$

I suspect that you're supposed to use this definition, though I can't quite tell for sure.

Given a set $EsubseteqBbb R,$ a point $x_0in E$ such that $(x_0-c,x_0+c)subseteq E$ for some $c>0,$ and a function $f:EtoBbb R,$ 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 $$f(x)-f(x_0)=Lcdot(x-x_0)+h(x)tag1$$ and $$lim_xto x_0frach(x)x-x_0=0.tag2$$ 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 $$f'(x_0)=frac12sqrtx_0$$ for each $x_0in(0,infty),$ in your case. Now, we need to show that $(1)$ and $(2)$ hold for some function $h$ when we substitute $L=frac12sqrtx_0.$ Fortunately, since we need $(1)$ to hold after substitution, it's easy to see that we require $$h(x)=f(x)-f(x_0)-Lcdot(x-x_0)=sqrtx-sqrtx_0-frac12sqrtx_0(x-x_0).$$ Then $$frach(x)x-x_0=fracsqrt x-sqrtx_0x-x_0-frac12sqrtx_0=frac1sqrt x +sqrtx_0-frac12sqrtx_0,$$ so to show that $(2)$ holds, it suffices to show that $$lim_xto x_0frac1sqrt x +sqrtx_0=frac12sqrtx_0.$$