<s>[INST] 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
$$f(x)=\dfrac{u(x)}{v(x)}\iff f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}.$$
 [/INST] 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.
</s>