If
\(\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{\mathrm{a}}{2}(x-|x|), & \text { for } x \lt 0 \ 0, & \text { for } x=0 \ b x^2 \end{array}\right.\) \(\sin \left(\frac{1}{x}\right), \text { for } x\gt0\) is continuous at \(x=0\), then

\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{f}(0)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \)
\( \therefore \lim _{x \rightarrow 0} \frac{\mathrm{a}}{2}(x-|x|)=0 \)
\( \therefore \lim _{x \rightarrow 0} \frac{\mathrm{a}}{2}[x-(-x)]=0 \ldots[\because x \lt 0 \Rightarrow|x|=-x] \)
\( \therefore \lim _{x \rightarrow 0} x=0\)
Which is true for any real value of a.
\(\lim _{x \rightarrow 0^{+}} f(x)=f(0)\)
\(\therefore \operatorname{limb}_{x \rightarrow 0} \mathrm{~b} x^2 \sin \left(\frac{1}{x}\right)=0\)
Note that \(x \neq 0\)
\(\therefore -1 \leq \sin \left(\frac{1}{x}\right) \leq 1\)
\(\therefore \) for any real value of b, above limit will be 0.