# Verification of Differentiability Proof

Let I be an open interval that contains 0 and

f: from I to R. If there exist an alpha greater than 1 such that

for all x, prove that f is differentiable at x=0.

What happens when alpha = 1?

Please check if my proof is correct and fix the graph.

Proof.

To prove that f(x) is differentiable, we need to prove that the following limit exists

By the limit theorems, the limit of the difference, is the difference of the limits.

In particular a=0, then is enough to check that that the limit of f(x)/x and

f(0)/x exist when x goes to 0.

By the comparison theorem for functions, the limit of the absolute value of x to the alpha goes to 0. Therefore, the limit of f(x) also goes to 0. Additionaly, for the squeeze theorem for functions f(0)=0. Therefore, the derivative exists and equals to 0.

If alpha is equal to 1, we have in the right side of the inequality the absolute value function that is continuous but not differentiable at x=0. As the derivative of the absolute value function is 1 when x is greater than 0, and -1 when x is less than 0, we would not be able to know if f(x) is differentiable at x=0.

• You've shown that $\lim_{x\to 0} f(x)=0$, but remember you were trying to show that $\lim_{x\to 0} \frac{f(x)}{x}$ exists. This isn't any harder, you just have to write it. (Depending on how thorough you're trying to be, you may want to explain why $f$ is continuous at zero when $\alpha=0$, and to give an example of functions which are/aren't differentiable there.) – Eric Stucky Nov 2 '16 at 23:50
• @EricStucky Thank you for your brilliant ideas! What about when alpha =1? – Beginner Nov 3 '16 at 1:13
• Erm, sorry, I meant $\alpha=1$ when I said $\alpha=0$. – Eric Stucky Nov 3 '16 at 14:35

(1). $|f(0)|\leq |0|^a=0$ so $f(0)=0.$
(2).For $x\ne 0$ we have $$\left| \frac {f(x)-f(0)}{x-0}\right|=\left|\frac {f(x)}{x}\right|=\frac {|f(x)|}{|x|}\leq \frac {|x|^a}{ |x|}=|x|^{a-1}.$$ And $\lim_{x\to 0}|x|^{a-1}=0$ for $a>1.$ So $f'(0)=0.$
• When $\alpha=1$ we cannot tell. E.g. if $f_1(x)=x$ for all $x$ then $f_1$ is differentiable everywhere. But if $f_2(x)=x\sin (1/x)$ for $x\ne 0,$ and $f_2(0)=0,$ then for $n\in \mathbb N$ we have : $f_2(x)/x=1$ when $x=1/((2n+\frac {1}{2}) \pi),$ but$f_2(x)/x=0$ when $x=1/(2n\pi).$ So $f_2'(0)$ does not exist. – DanielWainfleet Nov 3 '16 at 1:25