Differentiability of $x^{2/3}$ at $x=0$ (cusp)

Since $f(0) = 0$, the difference quotient is $\frac{f(h)-f(0)}{h} = \frac{h^{2/3}}{h} = h^{-1/3} = \frac{1}{h^{1/3}}$. As $h \to 0^+$, $h^{1/3} \to 0^+$, so $\frac{1}{h^{1/3}} \to +\infty$. As $h \to 0^-$, $h^{1/3} \to 0^-$, so $\frac{1}{h^{1/3}} \to -\infty$.

Since the one-sided limits are $+\infty$ and $-\infty$ respectively, the two-sided limit $\lim_{h\to 0} h^{-1/3}$ does not exist (not even as $\pm\infty$), because the two sides disagree in sign. This corresponds graphically to a cusp at $x=0$, where the tangent lines from the left and right both become vertical but point in opposite directions. Therefore $f'(0)$ does not exist, and $f(x)=x^{2/3}$ is not differentiable at $x=0$, even though it is continuous there.

The two one-sided limits approach opposite infinities. From the right the difference quotient tends to positive infinity, while from the left it tends to negative infinity. Because the signs disagree, the two-sided limit does not exist. This mismatch is exactly what produces the sharp cusp shape at the origin.