Differentiability at a cusp using the limit definition

Since $f(1) = 0$, the difference quotient simplifies to $\frac{f(x)-f(1)}{x-1} = \frac{(x-1)^{2/5}}{x-1} = (x-1)^{-3/5} = \frac{1}{(x-1)^{3/5}}.$ We now examine the limit of this quantity as $x \to 1$.

As $x \to 1^+$, we have $x - 1 > 0$, so $(x-1)^{3/5} \to 0^+$, and thus $\frac{1}{(x-1)^{3/5}} \to +\infty$. As $x \to 1^-$, we have $x - 1 < 0$, and the expression $(x-1)^{3/5}$ is not defined in the real number system for negative bases and non-integer exponents. Therefore, the left-sided limit does not exist. Since the right-sided limit is $+\infty$ (infinite) and the left-sided limit does not exist in the real numbers, the two-sided limit does not exist, so $f$ is not differentiable at $x=1$. Graphically, the slope of the secant line shoots off to $+\infty$ as we approach from the right, creating a vertical tangent line pointing upward. The function has a cusp at $x=1$, where the curve has a sharp corner with an infinite slope from the right.