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 ?
 A: 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.
A: 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.
A: 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).
A: 
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A: 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
A: 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.
