Differentiability of a function satisfying a condition on a bounded interval

Question

Let $\displaystyle f(x)$ satisfies the condition $|f(x)|\leq 1-\cos x$ on the interval $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$.

Prove that $f(x)$ is differentiable at $x=0$ and find $f^\prime(0)$.

My idea:

$\lim_{x\to 0} |\frac{f(x)-f(0)}{x}|\leq \lim_{x\to 0} \frac{|f(x)|+|f(0)|}{x}\leq \lim_{x\to 0}\frac{1-\cos x }{x}\leq \lim_{x\to 0}\frac{2 }{x}$

but i m stack in here

What have you tried so far? Have you written out the definition of $f'(0)$ in terms of a limit? — John Hughes, Dec 23 '14 at 13:11
I wrote but i still cant have any idea how can i proceed in the proof. — Raio, Dec 23 '14 at 13:41
Why don't you edit your question to describe what you've done so far? Can you tell us the value of $f(0)$? — John Hughes, Dec 23 '14 at 13:44
Write down the definition of the derivative of $\cos$ in $0$. (Why) does that help you? — PhoemueX, Dec 23 '14 at 14:06

score 3 · Answer 1 · edited Apr 13 '17 at 12:20

3

I think you are stuck because you are not familiar with the fact that $$\lim_{x\to 0}\frac{1-\cos x }{x}=0.$$

You can find the proof here: Evaluating $\lim_{x\to0}\frac{1-\cos(x)}{x}$

edited Apr 13 '17 at 12:20

Community

1

answered Dec 23 '14 at 17:14

Ludolila

3,034

score 2 · Accepted Answer · answered Dec 23 '14 at 14:25

2

Your condition forces $f(0)=0$. Now, $$ \frac{f(x)-f(0)}{x-0} = \frac{f(x)}{x}, $$ and $$ \left| \frac{f(x)}{x} \right| \leq \frac{1-\cos x}{|x|}. $$ Letting $x \to 0$, what do you get?

answered Dec 23 '14 at 14:25

Siminore

35,136

Differentiability of a function satisfying a condition on a bounded interval

2 Answers2