GATE CS 2007 Q1 — Calculus: Limits, Continuity, Differentiability

Let us analyze the two statements for the absolute value function $f(x) = |x|$:

The function can be written piecewise as:

1. Continuity (Statement P)

For any $x \neq 0$, $f(x)$ is a linear polynomial, which is continuous everywhere. 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 $x = 0$. Thus, $f(x)$ is continuous for all real values of $x$. Statement P is true.

2. Differentiability (Statement Q)

Let us check the differentiability of $f(x)$ at the critical point $x = 0$:

• Left-Hand Derivative (LHD) at $x = 0$: $$\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1$$
• Right-Hand Derivative (RHD) at $x = 0$: $$\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1$$

Since $\text{LHD} \neq \text{RHD}$ at $x = 0$, the derivative does not exist at $x = 0$. Thus, $f(x)$ is not differentiable for all real values of $x$. Statement Q is false.

Hence, P is true and Q is false.