Question

Prove that the function $f(x) = |x|$ is continuous but not differentiable at $x = 0$.

Hint:

Remember: Sharp corner (cusp) → continuous but not differentiable.

Answer 1

Concept: A function is continuous at a point if left-hand limit (LHL), right-hand limit (RHL), and function value are equal. It is differentiable if left-hand derivative (LHD) equals right-hand derivative (RHD). Step 1: {Definition of $|x|$.}
\[ |x| = \begin{cases} x, & x \ge 0 \\ -x, & x<0 \end{cases} \]
Step 2: {Check continuity at $x=0$.}
LHL: \[ \lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0 \] RHL: \[ \lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0 \] Function value: \[ f(0) = |0| = 0 \] Since LHL = RHL = $f(0)$, the function is continuous at $x=0$.
Step 3: {Check differentiability at $x=0$.}
LHD: \[ \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \] RHD: \[ \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]
Step 4: {Compare derivatives.}
\[ \text{LHD} = -1 \neq 1 = \text{RHD} \]
Step 5: {Conclusion.}
Since LHD $\ne$ RHD, the function is not differentiable at $x=0$. \[ \boxed{\text{$f(x) = |x|$ is continuous but not differentiable at } x=0} \]