Counterexample on the limit of $\frac{f(x)}{x}$

Question

Is the following statement true or false?

$f$ is defined on the set of all real numbers, such that $\lim \limits_{x\to 0} \dfrac{f(x)}{x}$ is a real number $L$ and $f(0)=0$. Then $L=0$?

I can't draw up any counterexample.

Would be grateful for hint.

You can't daw up $f(x)=x$? – Hagen von Eitzen Oct 23 '17 at 10:01 — Hagen von Eitzen, Oct 23 '17 at 10:01
Ohh yes. I have mistaken during the test. – RFZ Oct 23 '17 at 10:01 — RFZ, Oct 23 '17 at 10:01

score 2 · Answer 1 · answered Oct 23 '17 at 10:01

2

Simply set $f(x)=x$. Then $L=1$.

answered Oct 23 '17 at 10:01

Martin

score 2 · Answer 2 · answered Oct 23 '17 at 10:04

2

Take $f=\sinh(x)$ while $L=1$ and $f(0)=0$

answered Oct 23 '17 at 10:04

Mikasa

score 1 · Answer 3 · answered Oct 23 '17 at 10:02

1

Hint: Simply let $f(x)=x^2+x$

answered Oct 23 '17 at 10:02

mxaxc

score 1 · Answer 4 · answered Oct 23 '17 at 10:05

1

Infact $L$ can be any real number because $$\lim\limits_{x\to 0}\frac{\sin ax}x=a$$

answered Oct 23 '17 at 10:05

Dontknowanything

4 Answers4