If $f$ is a function differentiable at $a$ find: $\underset{h\rightarrow 0}{\lim} \frac{f(a+7h)-f(a-9h^2)}{h}$

Question

I am struggling to understand what to do. I tried brute forcing this question and I get $\frac{f(a)-f(a)}{0}$ which makes no sense to me. Any clarification would be appreciated.

score 2 · Answer 1 · answered Dec 05 '20 at 22:17

2

Hint: $\dfrac{f(a+7h)-f(a-9h^2)}{h} = \dfrac{(f(a+7h) -f(a)) + (f(a) - f(a - 9h^2))}{h} = 7\cdot\dfrac{f(a+7h) - f(a)}{7h} + 9h\cdot \dfrac{f(a) - f(a - 9h^2)}{9h^2}$

answered Dec 05 '20 at 22:17

Those steps make sense to me. However, I am struggling with 2 things. 1, I don't have a grasp with what exactly I am trying to solve and how an answer would even remotely look like. 2. I sense I am missing some definition or rule. By any change does $\dfrac{f(a+7h) - f(a)}{7h}$ equal $7f^{'}(x)$ or something? – Nuemann12 Dec 06 '20 at 04:27

score 0 · Answer 2 · answered Dec 06 '20 at 03:42

Why not to try with Taylor series ?

$$f(a+7h)=f(a)+7 h f'(a)+\frac{49}{2} h^2 f''(a)+O\left(h^3\right)$$ $$f(a-9h^2)=f(a)-9 h^2 f'(a)+O\left(h^3\right)$$ $$f(a+7h)-f(a-9h^2)=7 h f'(a)+h^2 \left(\frac{49 f''(a)}{2}+9 f'(a)\right)+O\left(h^3\right)$$ $$\frac{f(a+7h)-f(a-9h^2)}h=7 f'(a)+h \left(\frac{49 f''(a)}{2}+9 f'(a)\right)+O\left(h^3\right)$$ and $h \to 0$.

If $f$ is a function differentiable at $a$ find: $\underset{h\rightarrow 0}{\lim} \frac{f(a+7h)-f(a-9h^2)}{h}$

2 Answers2