Question

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

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

Hint:

For continuous functions, ensure that the limit at the point of interest matches the function's value at that point.

Answer 1

The Correct Option is A

For the function to be continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \) approaches 0 must equal the value of the function at \( x = 0 \). Therefore, we need to evaluate the limit: \[ \lim_{x \to 0} \frac{\sin^2(ax)}{x^2}. \] We know that \( \sin(x) \approx x \) as \( x \to 0 \), so \( \sin(ax) \approx ax \). Therefore: \[ \sin^2(ax) \approx a^2 x^2. \] Thus, the limit becomes: \[ \lim_{x \to 0} \frac{a^2 x^2}{x^2} = a^2. \] For the function to be continuous at \( x = 0 \), this limit must equal the value of \( f(0) = 1 \). Therefore: \[ a^2 = 1, \quad a = \pm 1. \] So, the correct value of \( a \) is \( \pm 1 \).