Find the value(s) of '$\lambda$', if the function $f(x) = \begin{cases} \frac{\sin^2 \lambda x}{x^2}, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \text{ is continuous at } x = 0.$

$\lambda = \pm 1$

The correct answer is Option (2) → $\lambda = \pm 1$ ##

$f(x) = \begin{cases} \frac{\sin^2 \lambda x}{x^2}, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$.

$f(0) = 1$

$\text{LHL} = \lim\limits_{x \to 0^-} \frac{\sin^2 \lambda x}{x^2}$

$= \lim\limits_{h \to 0} \frac{\sin^2 \lambda(0 - h)}{(0 - h)^2}$

$= \lim\limits_{h \to 0} \frac{\sin^2 \lambda h}{h^2}$

$= \lim\limits_{h \to 0} \lambda^2 \left( \frac{\sin \lambda h}{\lambda h} \right)^2$

$= \lambda^2 \times 1 = \lambda^2$

$f(0)=\lim\limits_{x \to 0^-}f(x)$

$1=\lambda^2$

$∴\lambda=\pm 1$