4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Hint: write the limit definition of the derivative "upside down." $\endgroup$ – Sean Roberson Jan 15 at 5:51
  • $\begingroup$ I tried it but I can't see how it helps me. $\endgroup$ – Igor Jan 15 at 5:53
14
$\begingroup$

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

$$\begin{equation}\begin{aligned} \lim_{h\rightarrow0}\frac{f(a+3h)-f(a-2h)}{f(a-5h)-f(a-h)} & = \lim_{h\rightarrow0}\frac{(f(a+3h)-f(a))-(f(a-2h)-f(a))}{(f(a-5h)-f(a))-(f(a-h)-f(a))} \\ & = \lim_{h\rightarrow0}\frac{\frac{f(a+3h)-f(a)}{h}-\frac{f(a-2h)-f(a)}{h}}{\frac{f(a-5h)-f(a)}{h}-\frac{f(a-h)-f(a)}{h}} \\ & = \lim_{h\rightarrow0}\frac{3\left(\frac{f(a+3h)-f(a)}{3h}\right)-(-2)\left(\frac{f(a-2h)-f(a)}{-2h}\right)}{(-5)\left(\frac{f(a-5h)-f(a)}{-5h}\right)-(-1)\left(\frac{f(a-h)-f(a)}{-h}\right)} \\ & = \frac{3f'(a) - (-2)f'(a)}{(-5)f'(a)-(-1)f'(a)} \\ & = \frac{5f'(a)}{(-4)f'(a)} \\ & = -\frac{5}{4} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

$\endgroup$
  • $\begingroup$ +1. Simple and clear. $\endgroup$ – Paramanand Singh Jan 15 at 6:42
9
$\begingroup$

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}

$\endgroup$
6
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.