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 ensure the function \( f(x) = \begin{cases} \frac{\sin^2 ax}{x^2}, & \text{if } x eq 0 \\ 1, & \text{if } x = 0 \end{cases} \) is continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) = 1 \).

First, we evaluate the limit: \(\lim_{x \to 0} \frac{\sin^2 ax}{x^2}\).

Using the known limit \(\lim_{y \to 0} \frac{\sin y}{y} = 1\), we substitute \( y = ax \). As \( x \to 0 \), \( y = ax \) also approaches 0.

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

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

For continuity at \( x = 0 \), we set the limit equal to \( f(0) \):

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

Solving for 'a' gives \( a = \pm 1 \).

Based on the options, the correct values for 'a' that ensure continuity are \( a = \pm 1 \).