0

I'm trying to solve an excercise. If I prove the equation below, I will be able to finish the rest of the exercise pretty quickly.

We know that: $$ \lim_{x \to 0} f(x) = 0 $$ $$ \lim_{x \to 0} \frac{f(2x)-f(x)}{x} = 0 $$

Can we then conclude (maybe from the definition of a derivative) that $f'(0)$ exists and is equal $0$?

CasseroleBoi
  • 31
  • 3
  • The result holds if $f$ is continuous at $0$. Otherwise derivative at $0$ does not exist. For the case when $f$ is continuous see https://math.stackexchange.com/a/1864092/72031 – Paramanand Singh Dec 21 '20 at 09:12
  • In the current case we can prove in similar manner that $f(x) /x\to 0$. See https://math.stackexchange.com/a/568871/72031 – Paramanand Singh Dec 21 '20 at 09:13

1 Answers1

0

Hint: If $f$ is differentiable at $0$, we need $f$ to be continuous at $0$, i.e. $\displaystyle\lim_{x \to 0}f(x) = f(0)$. However, we're only given $\displaystyle\lim_{x \to 0}f(x) = 0$. What happens if $f(0) \neq 0$? See if you can construct a function which satisfies $\displaystyle\lim_{x \to 0}f(x) = 0$ and $\displaystyle\lim_{x \to 0}\dfrac{f(2x)-f(x)}{x} = 0$, but $f(0) \neq 0$.

JimmyK4542
  • 54,331
  • Just to be sure. If $\lim_{x \to 0} f(x)$ exists and is finite, $f$ must be continuous at zero and attain that value, right? – ViktorStein Dec 21 '20 at 17:44