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# Find the value of 'a ' so that the function f(x) is defined by $f(x) = \begin{cases} \frac{\sin ^2 a x}{x^2} & x \neq 0 \\ 1 , & x =0 \end{cases}$ may be continuous at x = 0

$(A)\; a =\pm1$
$(B)\; a =0$
$(C)\; a =-1$
$(D)\; a =1$

$\lim_{x \to \infty} \exp(-x) = 0$ $\lim_{x\to 0}$
$\lim_{x \to 0} f(x) = f(0)$
$\lim_{x \to 0} \frac{\sin^2 ax}{x^2} = 1$
$a^2 \lim_{x \to 0} (\frac{\sin ax}{ax})^2 = 1$
But $\lim_{x \to 0} \frac{\sin \theta}{\theta} = 1$
$\therefore a^2 \times 1 = 1$
$\implies a =\pm1$