$$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$.
I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient?
Assume toward contradiction that $f$ is differentiable at $0$.
Then by definition we know that $f$ is continuous at $0$. Therefore
$\lim _{x\to 0}f\left(x\right)=f\left(0\right)$
We know that
$\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$
Therefore we can conclude
$\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$
But we can also rewrite this as
$\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$
Therefore $f(0) = 0$.
Since $f$ is differentiable at $0$ we have
$\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$
Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have
$\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$
$\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$
Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$.
Therefore $f$ is not differentiable at 0.
Best Answer
Your proof looks almost perfect to me. There's a couple of minor things I would change in the last few lines. You wrote:
I don't see the need to remove the absolute value signs in the following places in red:
$f'(0)= \lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} = \lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{\color{red}{|}x\color{red}{|}}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$, but
$f'(0) =\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} = \lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{ \color{red}{|}x\color{red}{|}} \cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$
Since $\lim \:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}$ only exists if both $\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}$ and $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}$ exist and are equal - which they aren't - we have a contradiction to the fact $f$ is differentiable at $0$.
Therefore $f$ is not differentiable at $0$.