Sometimes one defines $e$ as the (unique) number for which $$\tag 1 \lim_{h\to 0}\frac{e^h-1}{h}=1$$
In fact, there are two possible directions.
$(i)$ Start with the logarithm. You'll find out it is continuous monotone increasing on $\Bbb R_{>0}$, and it's range is $\Bbb R$. It follows $\log x=1$ for some $x$. We define this (unique) $x$ to be $e$. Some elementary properties will pop up, and one will be $$\tag 2 \lim\limits_{x\to 0}\frac{\log(1+x)}{x}=1$$
Upon defining $\exp x$ as the inverse of the logarithm, and after some rules, we will get to defining exponentiation of $a>0\in \Bbb R$ as $$a^x:=\exp(x\log a)$$
In said case, $e^x=\exp(x)$, as we expected. $(1)$ will then be an immediate consequence of $(2)$.
$(ii)$ We might define $$e=\sum_{k=0}^\infty \frac 1 {k!}$$ (or the equivalent Bernoulli limit). Then, we may define $$\exp x=\sum_{k=0}^\infty \frac{x^k}{k!}$$ Note $$\tag 3 \exp 1=e$$
We define the $\log$ as the inverse of the exponential function. We may derive certain properties of $\exp x$. The most important ones would be $$\exp(x+y)=\exp x\exp y$$ $$\exp'=\exp$$ $$\exp 0 =1$$
In particular, we have that $\log e=1$ by. We might then define general exponentiation yet again by $$a^x:=\exp(x\log a)$$
Note then that again $e^x=\exp x$. We can prove $(1)$ easily recurring to the series expansion we used.
ADD As for the definition of the logarithm, there are a few ones. One is $$\log x=\int_1^x \frac{dt}{t}$$
Having defined exponentiation of real numbers using rationals by $$a^x=\sup\{a^r:r\in\Bbb Q\wedge r<x\}$$
we might also define $$\log x=\lim_{k\to 0}\frac{x^k-1}{k}$$
In any case, you should be able to prove that
$$\tag 1 \log xy = \log x +\log y $$
$$\tag 2 \log x^a = a\log x $$
$$\tag 3 1-\dfrac 1 x\leq\log x \leq x-1 $$
$$\tag 4\lim\limits_{x\to 0}\dfrac{\log(1+x)}{x}=1 $$
$$\tag 5\dfrac{d}{dx}\log x = \dfrac 1 x$$
What you want is a direct consequence of either $(4)$ or $(5)$, or of the first sentence in my post.
ADD We can prove that for $x \geq 0$ $$\lim\left(1+\frac xn\right)^n=\exp x$$ from definition $(ii)$.
First, note that $${n\choose k}\frac 1{n^k}=\frac{1}{{k!}}\frac{{n\left( {n - 1} \right) \cdots \left( {n - k + 1} \right)}}{{{n^k}}} = \frac{1}{{k!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{k - 1}}{n}} \right)$$
Since all the factors to the rightmost are $\leq 1$, we can claim $${n\choose k}\frac{1}{{{n^k}}} \leqslant \frac{1}{{k!}}$$
It follows that $${\left( {1 + \frac{x}{n}} \right)^n}=\sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}} \leqslant \sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} $$
It follows that if the limit on the left exists, $$\lim {\left( {1 + \frac{x}{n}} \right)^n} \leqslant \lim \sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} = \exp x$$
Note that the sums in $$\sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}} $$
are always increasing, which means that for $m\leq n$
$$\sum\limits_{k = 0}^m {{n\choose k}\frac{{{x^k}}}{{{n^k}}}}\leq \sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}}$$
By letting $n\to\infty$, since $m$ is fixed on the left side, and $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{k!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{k - 1}}{n}} \right) = \frac{1}{{k!}}$$
we see that if the limit exists, then for each $m$, we have $$\sum\limits_{k = 0}^m {\frac{{{x^k}}}{{k!}}} \leqslant \lim {\left( {1 + \frac{x}{n}} \right)^n}$$
But then, taking $m\to\infty$ $$\exp x = \mathop {\lim }\limits_{m \to \infty } \sum\limits_{k = 0}^m {\frac{{{x^k}}}{{k!}}} \leqslant \lim {\left( {1 + \frac{x}{n}} \right)^n}$$
It follows that if the limit exists $$\eqalign{
& \exp x \leqslant \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n} \cr
& \exp x \geqslant \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n} \cr}$$ which means $$\exp x = \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n}$$ Can you show the limit exists?
The case $x<0$ follows now from $$\displaylines{
{\left( {1 - \frac{x}{n}} \right)^{ - n}} = {\left( {\frac{n}{{n - x}}} \right)^n} \cr
= {\left( {\frac{{n - x + x}}{{n - x}}} \right)^n} \cr
= {\left( {1 + \frac{x}{{n - x}}} \right)^n} \cr} $$
using the squeeze theorem with $\lfloor n-x\rfloor$, $\lceil n-x\rceil$, and the fact $x\to x^{-1}$ is continuous. We care only for terms $n>\lfloor x\rfloor$ to make the above meaningful.
NOTE If you're acquainted with $\limsup$ and $\liminf$; the above can be put differently as $$\eqalign{
& \exp x \leqslant \lim \inf {\left( {1 + \frac{x}{n}} \right)^n} \cr
& \exp x \geqslant \lim \sup {\left( {1 + \frac{x}{n}} \right)^n} \cr} $$ which means $$\lim \inf {\left( {1 + \frac{x}{n}} \right)^n} = \lim \sup {\left( {1 + \frac{x}{n}} \right)^n}$$ and proves the limit exists and is equal to $\exp x$.
Best Answer
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:
$$\begin{align} f'(x)&=\lim_{h\to0}\frac{\frac{5(x+h)+1}{2\sqrt{x+h}}-\frac{5x+1}{2\sqrt x}}h\\[1ex] &=\lim_{h\to0}\frac{\frac52\sqrt{x+h}+\frac1{2\sqrt{x+h}}-\frac52\sqrt x+\frac1{2\sqrt x}}h\\[1ex] &=\lim_{h\to0}\frac{\frac52\sqrt{x+h}-\frac52\sqrt x}h+\lim_{h\to0}\frac{\frac1{2\sqrt{x+h}}-\frac1{2\sqrt x}}h \end{align}$$