Kerala KEAM 5 June 2024 Solved Paper

We are given the following limit: $$\lim_{x \to 0} \frac{\sin \left( \pi \sin^2 x \right)}{x^2}.$$We need to evaluate this limit as $x \to 0$.Step 1: Use the approximation for small $x$, which is $\sin x \approx x$ when $x$ is close to 0. Therefore, for small $x$, we have: $$\sin^2 x \approx x^2.$$Thus, $\pi \sin^2 x \approx \pi x^2$.Step 2: Now, substitute $\pi x^2$ into the sine function: $$\sin \left( \pi \sin^2 x \right) \approx \sin \left( \pi x^2 \right).$$For small $x$, we can use the approximation $\sin y \approx y$ when $y$ is small. Therefore: $$\sin \left( \pi x^2 \right) \approx \pi x^2.$$Step 3: Substituting this approximation into the original limit expression, we get: $$\lim_{x \to 0} \frac{\sin \left( \pi x^2 \right)}{x^2} \approx \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi.$$Thus, the correct answer is option (E). Quick Tip: For small values of $x$, use approximations like $\sin x \approx x$ to simplify trigonometric expressions. These approximations help in evaluating limits.