Differentiability at a corner point

Since $f(2) = 0$, the difference quotient is $\frac{f(2+h) - f(2)}{h} = \frac{|h|}{h}$. For $h \to 0^+$, $|h| = h$, so this quotient equals $\frac{h}{h} = 1$, giving a right-hand limit of $1$.

For $h \to 0^-$, $|h| = -h$, so the quotient equals $\frac{-h}{h} = -1$, giving a left-hand limit of $-1$. Since the one-sided limits ($1$ and $-1$) do not agree, the overall limit $\lim_{h\to 0} \frac{|h|}{h}$ does not exist, so $f'(2)$ does not exist. Thus $f$ is not differentiable at $x = 2$, consistent with the sharp corner visible in the graph.

The right hand limit equals one, which is positive, while the left hand limit equals negative one, which is negative. This mismatch in sign is exactly why the two one-sided limits disagree. Because they disagree, the overall limit does not exist, confirming the corner at x equals two.