True or False: $f(x)$ is differentiable at $x=0$ and $\lim_{x \to 0}\frac{f(x)}{x}=3 \Rightarrow \; f(0)=0 $ and $f'(0)=3$? Prove or contradict:
$f(x)$ is differentiable at $x=0$ and $\lim_{x \to 0}\frac{f(x)}{x}=3 \Rightarrow \; f(0)=0 $ and $f'(0)=3$
My answer:
True and below the proof:
If $f(x)$ is differentiable at $0$ it means $f'(0)$ exists and verifies: $f'(0)=\lim_{x \to 0}\frac{f(x)-f(0)}{x}=\lim_{x \to 0}\frac{f(x)}{x}-\frac{f(0)}{x}$.
As by assumption it is given that $\lim_{x \to 0}\frac{f(x)}{x}$ exists and equal to $3$ and because $f(x)$ is differentiable at $x=0$ hence $f(x)$ is continuous at $x=0$ and so $f(0)$ exits it cames by limit arithmetic that: $f'(0)=\lim_{x \to 0}\frac{f(x)-f(0)}{x}=\lim_{x \to 0}\frac{f(x)}{x}-\frac{f(0)}{x}=\lim_{x \to 0}\frac{f(x)}{x}-\lim_{x \to 0}\frac{f(0)}{x}=3-\lim_{x \to 0}\frac{f(0)}{x}$
Now if $f(0) \neq 0 \Rightarrow \lim_{x \to 0}\frac{f(0)}{x} = +- \infty$ and so in such case $f'(0)$ will not exist. That's why $f(0)=0$ and if $f(0)=0$ it cames naturally that $f'(0)=3$.
Q.E.D
Is it correct?

If $f$ is differentiable at $0,$ then $f$ is continuous at $0.$
Indeed
$$\lim_{x\to 0}[f(x)-f(0)]=\lim_{x\to 0}{f(x)-f(0)\over x}x=f'(0)\cdot 0=0$$
Then
$$f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}{f(x)\over x}x= 3\cdot 0=0$$
Thus $$f'(0)=\lim_{x\to 0}{f(x)-f(0)\over x}=\lim_{x\to 0}{f(x)\over x}=3$$
