Limit calculation using derivative I encountered this exercise:
Let $f(x)$ be a differentiable function, and suppose  that there exists some $a$ where $f'(a) \ne 0 $.
Calculate the limit:
$$ \lim_{h\rightarrow0}\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)}. $$
I have no clue how I can solve this. I was trying to separate into two terms, and multiply and divide by $h$, but it solves just the numerator limit. What can be done with the denominator limit?
 A: A bit late this answer but I think it is worth mentioning it.
Since $f$ is differentiable, we know that


*

*$f(a+h) = f(a) + f'(a)h + o(h)$.


Now, replace $h$ by $3h,-2h,-5h,$ and $-h$ correspondingly and noting that $o(ch) = o(h)$ for any constant $c$ you get
$$\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)}= \frac{f(a) + 3hf'(a)+ o(h) - (f(a) - 2hf'(a) + o(h))}{f(a)-5hf'(a) + o(h)-(f(a) - hf'(a) + o(h))}$$
$$= \frac{5hf'(a)+o(h)}{-4hf'(a) + o(h)}\stackrel{h\to 0}{\longrightarrow}-\frac 54$$
A: For any $j \neq 0$, $(a + jh) - a = jh$ and with the transformation $k = jh$, you have
$$\lim_{h \to 0}\frac{f(a+jh) - f(a)}{jh} = \lim_{k \to 0}\frac{f(a + k) - f(a)}{k} = f'(a) \tag{1}\label{eq1A}$$
Thus, you get
$$\begin{equation}\begin{aligned}
\lim_{h\rightarrow0}\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)} & = \lim_{h\rightarrow0}\frac{(f(a+3h)-f(a))-(f(a-2h)-f(a))}{(f(a-5h)-f(a))-(f(a-h)-f(a))} \\
& = \lim_{h\rightarrow0}\frac{\frac{f(a+3h)-f(a)}{h}-\frac{f(a-2h)-f(a)}{h}}{\frac{f(a-5h)-f(a)}{h}-\frac{f(a-h)-f(a)}{h}} \\
& = \lim_{h\rightarrow0}\frac{3\left(\frac{f(a+3h)-f(a)}{3h}\right)-(-2)\left(\frac{f(a-2h)-f(a)}{-2h}\right)}{(-5)\left(\frac{f(a-5h)-f(a)}{-5h}\right)-(-1)\left(\frac{f(a-h)-f(a)}{-h}\right)} \\
& = \frac{3f'(a) - (-2)f'(a)}{(-5)f'(a)-(-1)f'(a)} \\
& = \frac{5f'(a)}{(-4)f'(a)} \\
& = -\frac{5}{4}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
A: Without loss, $a=0$.

Lemma. Let $f$ be any function differentiable at $0$. Then as $h\to 0$, $$\frac{f(Ah) -f(Bh)}{h}  \to (A-B)f'(0).$$

Proof. 
\begin{align}
\frac{f(Ah) -f(Bh)}{h} = \frac{f(Ah) -f(0)}{h} - \frac{f(Bh) -f(0)}{h} \to (A-B)f'(0).
\end{align}
Therefore by the Lemma and the product and quotient rules of limits, 
\begin{align} 
\frac{f(3h) - f(-2h) }{f(-5h) - f(-h)} 
&=\frac{f(3h) - f(-2h) }{h}\cdot \frac h{f(-5h) - f(-h)} 
\\
&\to (3-(-2))f'(0)\cdot\frac1{(-5-(-1))f'(0)} = \frac{-5}4.
\end{align}
A: $$ \lim_{h\rightarrow0}\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)}=\lim_{h\rightarrow0}\dfrac{3\dfrac{f(a+3h)-f(a)}{3h}+2\dfrac{f(a-2h)-f(a)}{-2h}}{-5\dfrac{f(a-5h)-f(a)}{-5h}+\dfrac{f(a-h)-f(a)}{-h}}=\frac{3f'(0)+2f'(0)}{-5f'(0)+f'(0)}.$$
