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?

• Hint: write the limit definition of the derivative "upside down." – Sean Roberson Jan 15 at 5:51
• I tried it but I can't see how it helps me. – Igor Jan 15 at 5:53

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{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}\tag{2}\label{eq2A}

• +1. Simple and clear. – Paramanand Singh Jan 15 at 6:42

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 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

$$\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)}= \frac{f(a) + 3hf'(a)+ o(h) - (f(a) - 2hf'(a) + o(h))}{f(a)-5hf'(a) + o(h)-(f(a) - hf'(a) + o(h))}$$ $$= \frac{5hf'(a)+o(h)}{-4hf'(a) + o(h)}\stackrel{h\to 0}{\longrightarrow}-\frac 54$$

$$\lim_{h\rightarrow0}\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)}=\lim_{h\rightarrow0}\dfrac{3\dfrac{f(a+3h)-f(a)}{3h}+2\dfrac{f(a-2h)-f(a)}{-2h}}{-5\dfrac{f(a-5h)-f(a)}{-5h}+\dfrac{f(a-h)-f(a)}{-h}}=\frac{3f'(0)+2f'(0)}{-5f'(0)+f'(0)}.$$