# Why is a function at sharp point not differentiable?

I am learning about differentiability of functions and came to know that a function at sharp point is not differentiable.
For eg. $$f(x)=|x|$$ I could find out that $f(x)$ is not differentiable at $x=0$ because
$$\lim_{x\to 0^-}f'(x) \ne \lim_{x\to 0^+}f'(x)$$ This is all mathematical but I couldn't understand where the sharp point plays its role here ?
How sharp point makes these limits to evaluate different ?

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Very informally, $f$ is differentiable at $a$ if a very very tiny bug sitting at $a$ can believe that the curve is flat at $a$, that its world is a straight line. The bug, sitting at $0$ on $y=|x|$, will not believe that the world is flat. Particularly if you turn the curve upside down, so that the "sharp" point is digging into its sitting end. – André Nicolas Jul 4 '12 at 8:25
+1: You have a very imaginative vision of the life of bugs , @André! – Georges Elencwajg Jul 4 '12 at 8:35
A variant of this always gets a laugh. And maybe they actually will remember. – André Nicolas Jul 4 '12 at 8:53

First remark: your $f$ is not differentiable (at $0$) because the limit $$\lim_{h \to 0} \frac{|h|}{h}$$ does not exist. In general the limit of $f'$ is only a sufficient condition for differentiability. Be very careful, if you use it to disprove differentiability.

Have you tried to sketch the graph of $f$? If so, you have seen that there is no tangent line to the graph at $0$, because of the sharp point. This is way to "understand" the rôle of the sharp point. But again, be careful: differentiability is a mathematical idea. The best way to understand it, is to understand it mathematically, according to the definition. Everything else may be misleading.

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For example, take $f(x)=x^2\sin(1/x)$. Then $\lim\limits_{x\to0}f'(x)$ does not exist, yet $f'(0)=0$. – robjohn Jul 4 '12 at 8:44
@robjohn But when we write $\lim_{x\to 0^-}f'(x) \ne \lim_{x\to 0^+}f'(x)$, I suppose we might be implying that the limits from the left and right actually exist. In this case, the inequality does in fact imply that the function is not differentiable at $0$. – Mark McClure May 19 '14 at 12:59
@MarkMcClure: I don't think that $\lim\limits_{x\to0}f(x)\ne0$ implies that $\lim\limits_{x\to0}f(x)$ exists, so I would really be hesitant to say that $\lim\limits_{x\to0}f(x)\ne\lim\limits_{x\to0}g(x)$ implies that either limit exists. – robjohn May 19 '14 at 14:24

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Note that all this seems to argue for is that there is no assignment of slope at the corner point that will make slope a continuous function of the input variable $x,$ not that it isn't possible to assign a value of the slope at the corner point. For instance, why couldn't the slope be $-1$ to the left of the corner point, $+1$ to the right of the corner point, and $0$ at the corner point? (Yes, I'm aware no assignment of slope is possible at a point where slope left of the point is $-1$ and slope right of the point is $+1,$ but the explanation is more real analysis than calculus.) – Dave L. Renfro Jul 5 '12 at 20:41
@DaveL.Renfro: Thanks for the comment, Dave. Here we see the left- and right-sided derivatives do not agree. This is robjohn's answer in visual form. – user26872 Jul 5 '12 at 23:06
It depends, if you define the absolute value as the limit of a "smooth" maximum, $|x|=\lim_{k\to\infty}\ln(e^{kx}+e^{-kx})/\ln k$ you might as well define $f'(x)=0$ (although real mathematicians will cringe at the sight of the reversed order of limits). However, without such crafty regularizations, the left and the right limits being different means that the limit does not exist. In terms of derivative meaning the slope of the curve, this answer is the most direct and comprehensible so far. +1. – orion May 19 '14 at 12:39

A function is differentiable at a point, $x_0$, if it can be approximated very close to $x_0$ by $f(x)=a_0+a_1(x-x_0)$. That is, up close, the function looks like a straight line. A kink, like you see in $|x|$ at $x=0$, is where the graph of $|x|$ does not look like a straight line.

Rather than look at $$\lim\limits_{h\to0^+}f'(x+h)\quad\text{and}\quad\lim\limits_{h\to0^-}f'(x+h)\tag{1}$$ w should look at $$\lim\limits_{h\to0^+}\frac{f(x+h)-f(x)}{h}\quad\text{and}\quad\lim\limits_{h\to0^-}\frac{f(x+h)-f(x)}{h}\tag{2}$$ If $f$ is continuous and the limits in $(1)$ exist and are equal, then $f'(x)$ is equal to those limits. However, if $$f(x)=x^2\sin(1/x)\tag{3}$$ then the limits in $(1)$ do not exist for $x=0$, yet $f'(0)=0$.

However, by definition, if and only if the limits in $(2)$ exist and are equal, does $f'(x)$ exist and equal to those limits.

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It's not differentiable because you can draw infinitely many tangents that touch the point of turning (I may be wrong I'm just a high school student).

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Because in order to calculate a derivative you need to take the difference of the point in question and the next point on the closest possible interval. This means that you must take the difference from the right and from the left. When you encounter a kink if you take the difference of the closest point from the left and from the right you will get reciprocal answers. Therefor it is impossible to determine the rate of change at the point in question

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could we say that as we approach from the left and right, the functions have not 'turned' towards each other e,g, to a point where slope will be zero ? Hence a turning point that is curved IS differentiable, but this 'cusp' is not.

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Though this may provide an intuitive explanation, it may not be accepted as an answer in a calculus paper. – Shailesh May 2 at 0:11