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I need to prove, or give a counter example:

If $\lim_{x \to 0} \dfrac{f(x)}{|x|} = 1$, then $f$ is not differentiable at 0.

I think it's true but can't formalize a proof (I tried proving it straight from the definition of the derivative. Thanks.

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If $f(0)$ is not defined, or different from $0$, then $f$ is not continuous at $0$, hence not differentiable there. Otherwise, look at $\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}$. By the limit assumption you were given, for $x$ positive and near enough to $0$, this is near $1$, while for $x$ negative and near enough to $0$, this is near $-1$, so the limit cannot exist. – André Nicolas Mar 3 '12 at 17:29
and then when x approaches 0 from left the lim is -1, and from right is 1... Thanks! – Amihai Zivan Mar 3 '12 at 17:33
@Amihai What the statement is saying is that, for $x$ near $0$, the function behaves in the same way as $|x|$. Is $|x|$ differentiable at $x=0$? – Pedro Tamaroff Mar 4 '12 at 5:55
up vote 4 down vote accepted

The way to think about this problem is that the condition you give means $f(x)$ looks $|x|$ (plus higher order terms) near $x = 0$. And since $|x|$ is not differentiable at $x = 0$, neither is $f(x)$.

To be more precise about it, use the fact that the limit is defined and this means that the two directional derivatives --- from the left and from the right --- must exist and are equal.

First consider the left limit, $x \leq 0$. $$ \lim_{x \nearrow 0} \frac{f(x)}{|x|} = \lim_{x \nearrow 0} \frac{f(x)}{-x} = -\lim_{x \nearrow 0} \frac{f(x)}{x} = 1 \Rightarrow \lim_{x \nearrow 0} \frac{f(x)}{x} = -1\,. $$

Now consider the right limit, $x \geq 0$. $$ \lim_{x \searrow 0} \frac{f(x)}{|x|} = \lim_{x \searrow 0} \frac{f(x)}{x} = 1 \Rightarrow \lim_{x \searrow 0} \frac{f(x)}{x} = 1\,. $$

Now consider the definition of the derivative of $f$ at $x$. $$ \frac{df}{dx}\!(0) = \lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\,. $$ This limit does not exist, because the two above directional limits disagree. Hence, $f$ is not differentiable at $0$.

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Note that $$\lim\limits_{h\rightarrow 0}f(h)= \lim\limits_{ h\rightarrow 0}|h|\cdot \lim\limits_{h\rightarrow0 }{f(h)\over |h|}=0\cdot1=0.$$

Assuming $f$ is continuous at $0$, we have $f(0)=0$.


$$\lim\limits_{h\rightarrow0^+}{f(h)\over h}=\lim\limits_{h\rightarrow0^+}{f(h)\over |h|}=1,$$


$$\lim\limits_{h\rightarrow0^-}{f(h)\over h}=-\lim\limits_{h\rightarrow0^-}{f(h)\over -h} =-\lim\limits_{h\rightarrow0^-}{ f(h)\over |h|}=-1;$$

whence $f'(0)=\lim\limits_{h\rightarrow0}{f(h)\over h}$ does not exist.

If $f$ is not continuous at 0, then of course $f$ is not differentiable at 0.

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