NIMCET 2020 — Mathematics PYQ

Detailed Step-by-Step Explanation:

To find the limit ${\lim }_{x\to 0}{x}^2{e}^{\sin \left(\frac{1}{x}\right)}$, we can use the Squeeze Theorem (also known as the Sandwich Theorem).

1. Analyze the bounded function:

   We know that the sine function is always bounded between $-1$ and $1$ for any real input. Therefore, for all $x\ne 0$:

   $$-1\le \sin \left(\frac{1}{x}\right)\le 1$$

2. Apply the exponential function:

   Since $e^y$ is a strictly increasing function, we can exponentiate all sides of the inequality without changing the direction of the inequality signs:

   $$e^{-1}\le e^{\sin \left(\frac{1}{x}\right)}\le e^1$$

3. Multiply by $x^2$:

   Since x^2 > 0 for all $x\ne 0$, multiplying the entire inequality by $x^2$ keeps the inequality signs intact:

   $$x^2{e}^{-1}\le x^2{e}^{\sin \left(\frac{1}{x}\right)}\le x^{2}e$$

4. Apply the limit as $x\to 0$:

   Now, we take the limit of the lower and upper bounds as $x$ approaches $0$:

   • Lower bound: ${\lim }_{x\to 0}{x}^2{e}^{-1}=0\cdot e^{-1}=0$

   • Upper bound: ${\lim }_{x\to 0}{x}^{2}e=0\cdot e=0$

Conclusion by Squeeze Theorem:

Since both the lower bound and the upper bound approach $0$, the function trapped between them must also approach $0$.

$$\underset{x\to 0}{\lim }{x}^2{e}^{\sin \left(\frac{1}{x}\right)}=0$$

Since $0$ is not explicitly given in options A, B, or C, the correct choice is D.