Mathematics - Limits Question with Solution | TestHub

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