If $f\left(x\right)=\left\{\begin{array}{l}x\sin \left(\frac{1}{x}\right),\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\ne 0\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=0\end{array}\right.$ Then $\underset{x\to 0}{lim} f\left(x\right)$

Solution:

We have,

$\begin{array}{l}f\left(x\right)=\left\{\begin{array}{c}x\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}\right.\\ \underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{x\to 0}{lim} f(0-h)=\underset{x\to 0}{lim} -h\sin \left(-\frac{1}{h}\right)\\ \Rightarrow \underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{x\to 0}{lim} h\sin \left(\frac{1}{h}\right)=0\end{array}$

Similarly, we obtain $\underset{x\to 0^{+}}{lim} f\left(x\right)=0.\text{ Hence, }\underset{x\to 0}{lim} f\left(x\right)=0$