GATE CS 1998 Q1.4 — Calculus: Limits, Continuity, Differentiability

To analyze the continuity and differentiability of the function $f(x) = |x|$ in the interval $[-1, 1]$, we examine its behavior at the critical point $x = 0$.

The function is defined as:

1. Continuity

For any point $x \neq 0$, $f(x)$ is a polynomial function, which is always continuous. At $x = 0$:

• Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$
• Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$
• Function value: $f(0) = 0$

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$, the function is continuous at $x = 0$, and thus continuous throughout the interval $[-1, 1]$.

2. Differentiability

We check the differentiability at $x = 0$ using the definition of derivatives:

• Left-hand derivative (LHD):$$\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$$
• Right-hand derivative (RHD):$$\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$$

Since $\text{LHD} \neq \text{RHD}$ at $x = 0$, the derivative does not exist at $x = 0$. Geometrically, the graph of $y = |x|$ has a sharp corner at $(0,0)$.

Therefore, the function is continuous but not differentiable in the interval $[-1, 1]$.