Prove or disprove: Let $f$ be a function that is defined on a neighbourhood of $0$. Suppose that $\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)}{\left|x\right|}=1$ , then $f$ is not differentiable at $0$ I somehow think its false but couldn't find a counter example. So I tried proving it and this is my prove of it being true: let us assume by contradiction, that $f$ is differentiable at $0$. thus: $\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)-f\left(0\right)}{x}=f'\left(0\right)$ Which implies: $f'\left(0\right)=\underset{x\rightarrow0^+}{\lim}\frac{f\left(x\right)-f\left(0\right)}{x}=\underset{x\rightarrow0^+}{\lim}\underbrace{\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]}$ and: $f'\left(0\right)=\underset{x\rightarrow0^-}{\lim}\frac{f\left(x\right)-f\left(0\right)}{x}=\underset{x\rightarrow0^-}{\lim}{\left[\frac{f\left(x\right)}{-\left|x\right|}-\frac{f\left(0\right)}{-\left|x\right|}\right]}=\underset{x\rightarrow0^-}{\lim}-\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]=\underbrace{-\underset{x\rightarrow0^+}{\lim}{\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]}}$ They must be equal, thus: $\underset{x\rightarrow0}{\lim}{\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]}=0$ As we know $\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)}{\left|x\right|}=1$, then: $\underset{x\rightarrow0}{\lim}\frac{f\left(0\right)}{\left|x\right|}=1$ which cant be, because if $f(0)=0$, then the limit is $0$, and if $f(0)\ne0$, the the lim is $\pm\infty$ Which in both cases, doesn't equal $1$, Contradiction. . . Was there a problem separating to $0\pm$ and change $x$ to $|x|$ without modifying $f(x)$? Or was there any other problem ? First post, be gentle. Thank you!
$$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ 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.
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