Limit of $\frac{f(1+h)-f(1-3h)}{5h}$ 
Assume that a function $f$ satisfies the condition $f'(1) = 2.$ Figure out the limit $$\lim_{h\to0}\frac{f(1+h)-f(1-3h)}{5h}.$$

This seemed to be a very simple problem just dealing with the definition of a derivative, but the proposed solution was a bit confusing. It went as follows:
Adding $-f(1)$ and $f(1)$ to the numerator we get
$$\frac{f(1+h)-f(1-3h)}{5h} = \frac{f(1+h)- f(1) +f(1)-f(1-3h)}{5h}. \tag{1}\label{eq1A}$$ 
This can be simplified to 
$$\frac{f(1+h)-f(1)}{5h} + \frac{f(1)-f(1-3h)}{5h}. \tag{2}\label{eq2A}$$
And now from here we get
$$\frac15 \cdot \frac{f(1+h)-f(1)}{h} + \frac35 \cdot \frac{f(1)-f(1-3h)}{3h}. \tag{3}\label{eq3A}$$
Furthermore
$$\frac15 \cdot \frac{f(1+h)-f(1)}{h} + \frac35 \cdot \frac{f(1-3h)-f(1)}{-3h}. \tag{4}\label{eq4A}$$
Now taking the limit $h \to 0$, we have that
$$\lim_{h\to0} \frac15 \cdot \frac{f(1+h)-f(1)}{h} + \lim_{h\to0} \frac35 \cdot \frac{f(1-3h)-f(1)}{-3h} = \frac15\cdot f'(1) + \frac35\cdot f'(1) = \frac85$$
Could someone educate me on parts $(3)$ and $(4)$. Why do we chose $\frac35$ in $(3)$ instead of $\frac15$? If we would have taken out $\frac15$ the result would have been $\frac15\cdot f'(1) + \frac15\cdot f'(1) = \frac45$ right? Also what's happening on $(4)$ how do we suddenly flip the numerator?
 A: You are assuming that$$\lim_{h\to0}\frac{f(1+h)-f(1)}h=2.$$This means that, for every number $a\in\mathbb R\setminus\{0\}$,$$\lim_{h\to0}\frac{f(1+ah)-f(1)}{ah}=2,\tag1$$too. Those manipulations were done so that you had to deal with limits of the form $(1)$.
A: Choosing $\frac{1}{5}$ instead of $\frac{3}{5}$ would not have worked because $$\lim_{h\to 0}\frac{f(1-3h) - f(1)}{-h} \neq f'(1)$$
A: Here's the basic idea. Define $a(h):=\frac{f(1+h)-f(1)}{h}$. Since $\displaystyle \lim_{h\to 0} a(h)=f'(1),$ the function $a(h)$ is continuous at $h=0$. Take any function $g(h)$ such that $\displaystyle \lim_{h\to 0} g(h)=0$. Then
$$
\lim_{h\to 0}a(h)=\lim_{h\to 0}a(g(h))=f'(1).
$$
In both difference quotients, the idea is to come up with an appropriate $g(h)$. For the first term, it's written as 
$$
\frac{f(1+h)-f(1)}{5h}.
$$
When $1/5$ gets pulled out, it becomes just the standard definition of a derivative so we leave it (i.e. $g(h)=h$)
For the second term,
$$
\frac{f(1)-f(1-3h)}{5h}=\frac{f(1-3h)-f(1)}{-5h},
$$
the most sensible substitution is to say $g(h)=-3h$ in an effort to replicate $f(1+g(h))-f(1)$, but we see that the denominator in the difference quotient is not $g(h)$. We want the denominator to be $g(h)=-3h$. Hence, pull out $1/5$ and multiply and divide by $3$. Then it becomes
$$
\frac{3}{5}\frac{f(1+g(h))-f(1)}{g(h)}.
$$
This will now evaluate to $\frac{3}{5}f'(1)$.
A: More generally,
since
$f(x+h)
\approx f(x)+hf'(x)+O(h^2)
$,
$\begin{array}\\
\dfrac{f(x+ah)-f(x+bh)}{h}
&=\dfrac{f(x)+ahf'(x)+O(h^2)-(f(x)+bhf'(x)+O(h^2))}{h}\\
&=\dfrac{(ah-bh)f'(x)+O(h^2)}{h}\\
&=(a-b)f'(x)+O(h)\\
\end{array}
$
Here
$x=1, a=1, b=-3$
so
the result is
$4f'(1)/5
=8/5
$.
