If _ x arrow x ((1)/(x))=A and _ x arrow 0 x ((1)/(x))=B , then...

As given, $\mathrm{A}=\lim _{\mathrm{x} \rightarrow \infty} \mathrm{x} \sin$\left(\frac{1}{\mathrm{x}}\right)=\lim _{\mathrm{x} \rightarrow \infty} \frac{\sin \left(\frac{1}{\mathrm{x}}\right)}{\left(\frac{1}{\mathrm{x}}\right)}$Let$\mathrm{t}=\frac{1}{\mathrm{x}}$when$\mathrm{x} \rightarrow \alpha, \mathrm{t} \rightarrow 0 $$\begin{aligned} \Rightarrow\mathrm{A}&=\lim _{t \rightarrow \infty} \frac{\sin \mathrm{t}}{\mathrm{t}}=1 \\ \text{ and } \left[\because \lim _{\mathrm{t} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=1\right] \end{aligned}$$ and \mathrm{B}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right) \Rightarrow \mathrm{B}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \cdot \lim _{\mathrm{x} \rightarrow 0} \sin \left(\frac{1}{\mathrm{x}}\right) \Rightarrow \mathrm{B}=0$\therefore \quad A=1 and B=0$ is correct