Question

If \( f(x) = \begin{cases} \frac{\sin^2 ax}{x^2}, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \) is continuous at \( x = 0 \), then the value of 'a' is :

• A. 1
• B. -1
• C. 0
• D. \( \pm 1 \)

Hint:

For continuity at a point, ensure that the left-hand limit and right-hand limit match the function's value at that point.

Answer 1

The Correct Option is D

To determine the value of 'a' that makes the function \( f(x) = \begin{cases} \frac{\sin^2 ax}{x^2}, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \) continuous at \( x = 0 \), we need to check the limit of \( f(x) \) as \( x \) approaches 0 and ensure it equals \( f(0) = 1 \).

First, find \(\lim_{x \to 0} \frac{\sin^2 ax}{x^2}\).

Using the well-known limit \(\lim_{y \to 0} \frac{\sin y}{y} = 1\), let \( y = ax \), then as \( x \to 0 \), \( y = ax \to 0 \) as well.

Thus, \(\lim_{x \to 0} \frac{\sin ax}{x} = \lim_{y \to 0} \frac{\sin y}{y} \cdot a = a\).

Therefore, \(\lim_{x \to 0} \frac{\sin^2 ax}{x^2} = \lim_{y \to 0} \left(\frac{\sin y}{y}\right)^2 \cdot a^2 = a^2\).

To ensure continuity at \( x = 0 \), we require:

\(\lim_{x \to 0} f(x) = f(0) \Rightarrow a^2 = 1\).

This yields \( a = \pm 1 \).

Given the options, the correct value of 'a' that aligns with the provided options is \( a = \pm 1 \).