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

Question

I figure that the answer is $\infty$, but I a torn on whether I am correct. Any idea whether I am correct or if I have an issue.

My work is the following: $\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$ = $\underset{h\rightarrow 0}{\lim} \frac{-(f(a-h^2)-f(a))}{h^2}$ = $\underset{h\rightarrow 0}{\lim} \frac{-1}{h}\frac{f(a-h^2)-f(a)}{h}$= $\infty \cdot f{'}(a)= \infty.$

edit: $\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$ = $\underset{h\rightarrow 0}{\lim} \frac{-h}{-h}\frac{f(a-h^2)-f(a)}{h}$= $\underset{h\rightarrow 0}{\lim} {-h}\frac{f(a-h^2)-f(a)}{-h^2}$= $0 \cdot f{'}(a)= 0.$

How you obtain first equality? And how title is connected to question limit? — zkutch, Dec 09 '20 at 03:30
You are looking at a different function than the one in the title — Andrei, Dec 09 '20 at 03:30
Your mistake is, that $h \to 0$, not $h \to \infty$ in first multiplier. — zkutch, Dec 09 '20 at 03:43

score 2 · Accepted Answer · answered Dec 09 '20 at 03:39

2

$$ \frac{f(a+h^2)-f(a)}{h}=h\frac{f(a+h^2)-f(a)}{h^2}\underset{h\rightarrow 0}{\longrightarrow}0 $$ Because $\lim\limits_{h\rightarrow 0}\frac{f(a+h^2)-f(a)}{h^2}=f'(a)$.

answered Dec 09 '20 at 03:39

Tuvasbien

8,907

I would explain that last line a bit more. It is false in general. It is true here because $f$ is differentiable, meaning the one-sided limit $\lim_{h \rightarrow 0^+} \frac{f(a+h) - f(a)}{h} = \lim_{h \rightarrow 0} \frac{f(a+h^2) - f(a)}{h^2}$ equals the two-sided limit $f'(a)$, but this bridge between one-sided and two-sided limits is missing. – Eric Towers Dec 09 '20 at 03:56

Eric Towers · Answer 2 · 2020-12-09T04:10:14.047

Since $f$ is continuous, $f(a-h^2) - f(a) \rightarrow 0$ as $h \rightarrow 0$, so we have a limit of the indeterminate form $\left[ \frac{0}{0} \right]$. Applying L'Hopital's rule, we obtain \begin{align*} \lim_{h \rightarrow 0} \frac{f(a-h^2) - f(a)}{h} &\overset{L'H}= \lim_{h \rightarrow 0} -2 h f'(a - h^2) \\ &= -2 \cdot 0 \cdot f'(a) \\ &= 0 \text{,} \end{align*} where in the last, we have used that differentiability of $f$ at $a$ implies continuity on an open set containing $a$.

score 1 · Answer 3 · answered Dec 09 '20 at 03:53

Your answer is correct. You can solve the problem by using L'hospital rule: $\displaystyle \lim_{h \to 0} \dfrac{f(a-h^2) - f(a)}{h}= \displaystyle \lim_{h \to 0} \dfrac{\frac{df(a-h^2)}{dh}}{1}= \displaystyle \lim_{h \to 0} (-2h\cdot f'(a))=0. $

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

3 Answers3