Question

Evaluate: $\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{\mathrm{\pi }}{x}\right)$

• A. $1$
• B. $0$
• C. Dose not exist
• D. $\infty$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Determine the value of $\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{\mathrm{\pi }}{x}\right)$

Consider the given Equation as,

$I=\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{\mathrm{\pi }}{x}\right)$

Using sandwich theorem

$$\begin{gathered} \underset{x\to 0}{\lim }-x^2\le \underset{x\to 0}{\lim }{x}^2\sin \left(\frac{\mathrm{\pi }}{x}\right)\le \underset{x\to 0}{\lim }{x}^2 \\ \Rightarrow 0\le I\le 0 \\ \Rightarrow I=0 \end{gathered}$$

[The squeeze (or sandwich) theorem states that if$f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for

all numbers, and at some point $x=k$ we have$f\left(k\right)=h\left(k\right),theng\left(k\right)$ must also be equal to them.]

Hence, the correct answer is option B.