Find the derivative using the definition of derivative (limit). 
Given $f(x)=\dfrac{5x+1}{2\sqrt{x}}$. Find $\dfrac{df(x)}{dx}=f'(x)$ using
the definition of derivative.

I have tried as below.
\begin{align*}
f'(x)&=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5(x+h)+1}{2\sqrt{x+h}}-\dfrac{5x+1}{2\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{\left(5(x+h)+1\right)\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x+h}-\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
\end{align*}
Now I can't find the limit. I confused how to simplify the limit. Anyone can give me  hint to solve it?
Note:
We were asked to find this derivative using
$$f'(x)=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$
instead of
$$f(x)=\dfrac{u(x)}{v(x)}\iff f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}.$$
 A: You can expand $f(x)$ as $\frac{5x+1}{2\sqrt x}=\frac52\sqrt x+\frac1{2\sqrt x}$, then differentiate the two.
$$\begin{align}
f'(x)&=\left(\frac52\sqrt x\right)'+\left(\frac1{2\sqrt x}\right)'\\[1ex]
&=\lim_{h\to0}\frac{\frac52\sqrt{x+h}-\frac52\sqrt x}h+\lim_{h\to0}\frac{\frac1{2\sqrt{x+h}}-\frac1{2\sqrt x}}h
\end{align}$$
But if you don't already know or cannot directly use the fact that differentiation distributes over sums, you can instead expand the limand in your second line and regroup the terms in the numerator, then separate the limit into two. You would end up arriving at the same point either way:
$$\begin{align}
f'(x)&=\lim_{h\to0}\frac{\frac{5(x+h)+1}{2\sqrt{x+h}}-\frac{5x+1}{2\sqrt x}}h\\[1ex]
&=\lim_{h\to0}\frac{\frac52\sqrt{x+h}+\frac1{2\sqrt{x+h}}-\frac52\sqrt x+\frac1{2\sqrt x}}h\\[1ex]
&=\lim_{h\to0}\frac{\frac52\sqrt{x+h}-\frac52\sqrt x}h+\lim_{h\to0}\frac{\frac1{2\sqrt{x+h}}-\frac1{2\sqrt x}}h
\end{align}$$
A: First, note that we don't need to worry about the derivative of $f$ when $x=0$, because your function isn't defined when $x=0$.
Using the Binomial expansion, as $h \to 0,$ which garuntees that $|\frac{h}{x}| < 1,$
$ \sqrt{x+h} = \sqrt{x} \sqrt{1 + \frac{h}{x}} = \sqrt{x} \left(1 + \frac{h}{x}\right)^\frac12 = \sqrt{x}\left(1 + \frac{\left(\frac12\right)}{1!\ } \left(\frac{h}{x}\right) + \frac{\left(\frac12\right)\left(\frac{-1}{2}\right) }{2!\ } \left(\frac{h}{x}\right)^2 + ...\right),$
i.e., $\ \sqrt{x+h} = \sqrt{x} + \frac{h}{2\ \sqrt{x} } + O(h^2).$
Substitute this into the last line of working in the question (but not the denominator), we get:
\begin{align*}
f'(x)&= \lim\limits_{h\to 0}
  \left(\dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\right)\\
&= \lim\limits_{h\to 0}
  \left(\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-(5x+1)\left(\sqrt{x}+ \frac{h}{2\ \sqrt{x} } + O(h^2)\right)}{2h\sqrt{x+h}\sqrt{x}}\right)\\
&= \lim\limits_{h\to 0}
  \left(\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x}-\frac52 h \sqrt{x} - \sqrt{x} - \frac{h}{2\ \sqrt{x} } }{2h\sqrt{x+h}\sqrt{x}} + O(\sqrt{h})\right).\\
\end{align*}
... and after some cancellation you will arrive at the answer.
A: This is not going to be pretty, but it was not intended to so here goes:
To make it managable, let us define
$$
\begin{align}
a&:=5(x+h)+1\\
b&:=\sqrt{x+h}\\
c&:=5x+1\\
d&:=\sqrt x
\end{align}
$$
Then we have
$$
\begin{align}
f(x+h)-f(x)&=\frac a{2b}-\frac c{2d}\\
&=\frac{ad-cb}{2bd}\\
&=\frac{(ad-cb)(ad+cb)}{2bd(ad+cb)}\\
&=\frac{a^2d^2-c^2 b^2}{2bd(ad+cb)}
\end{align}
$$
where
$$
\begin{align}
a^2d^2-c^2 b^2&=(5h+(5x+1))^2x-(5x+1)^2(x+h)\\
&=5h(5h+2(5x+1))x-(5x+1)^2h
\end{align}
$$
or something like that (might need to double check). Hence you can divide the numerator by $h$ and both limits of numerator and denominator are non-zero and can be determined. Thus the limit of quotient will follow from that. Pretty tedious expressions!
