NIMCET 2020 — Mathematics PYQ

Topic: Limits (Squeeze Theorem)

Question

What is the value of $\underset{x\to 0}{\lim }{x}^2{e}^{\sin \left(\frac{1}{x}\right)}$?

Step-by-Step Solution

1. Analyze the Trigonometric Function:

The function $\sin \left(\frac{1}{x}\right)$ is an oscillating function. Regardless of the value of $x$ (as long as $x\ne 0$), the value of sine always stays within the range:

2. Analyze the Exponential Function:

Since the exponential function $e^u$ is strictly increasing, we can apply the bounds from the sine function to the exponent:

3. Multiply by $x^2$:

Since $x^2$ is always non-negative ($x^2\ge 0$), the direction of the inequalities remains the same when we multiply:

4. Apply the Squeeze Theorem (Sandwich Theorem):

Now, we find the limits of the bounding functions as $x\to 0$:

• Lower Bound: $\underset{x\to 0}{\lim }\frac{x^2}{e}=\frac{0^2}{e}=0$

• Upper Bound: $\underset{x\to 0}{\lim }e{x}^2=e(0^2)=0$

5. Conclusion:

Since both the lower and upper bounds approach $0$, the function in the middle must also approach $0$.

Final Answer:

The value of the limit is: