Question

Evaluate $\underset{x\to \infty }{\lim }x\sin \left(\frac{2}{x}\right)$

• A. $\infty$
• B. $0$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Evaluating the Limit $\underset{x\to \infty }{\lim }x\sin \left(\frac{2}{x}\right)$

When we substitute limit, it becomes the value $\infty$ does not exist

$$\begin{gathered} =\underset{x\to \infty }{\lim }x\sin \left(\frac{2}{x}\right) \\ =\underset{x\to \infty }{\lim }\frac{x\sin \left(\frac{2}{x}\right){\displaystyle \frac{2}{x}}}{{\displaystyle \frac{2}{x}}}\mathbf{}\mathbf{[}\mathbf{\text{multiply and divide by}}\mathbf{}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{]} \\ =\underset{x\to \infty }{\lim }\frac{2}{x}.x.\frac{\sin \left(\frac{2}{x}\right)}{{\displaystyle \frac{2}{x}}} \\ =\underset{x\to \infty }{\lim }2.\frac{\sin \left(\frac{2}{x}\right)}{{\displaystyle \frac{2}{x}}} \end{gathered}$$

Let us assume $y=\frac{2}{x}$

Where $x\to \infty \Rightarrow y\to \frac{2}{\infty }=0$

$$\begin{gathered} =\underset{x\to \infty }{\lim }2\times \underset{y\to 0}{\lim }\frac{\sin \left(y\right)}{{\displaystyle y}}\mathbf{[}\mathbf{\because }\mathbf{l}\mathbf{i}\mathbf{m}\mathbf{i}{\mathbf{t}}_{\mathbf{x}\mathbf{\to }\mathbf{0}}\frac{\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{{\displaystyle x}}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{]} \\ =2\times 1 \\ =2 \end{gathered}$$

Hence, option (C) is the correct answer.