$\operatorname{Lt}_{x \rightarrow-\infty}\left[\frac{\left(x^{4} \sin \left(\frac{1}{x}\right)+x^{2}

Question

$\operatorname{Lt}_{x \rightarrow-\infty}\left[\frac{\left(x^{4} \sin \left(\frac{1}{x}\right)+x^{2}\right)}{\left(1+|x|^{3}\right)}\right]=$

Professor PC Thomas Classes · Answer

$\underset{x ightarrow-\infty}{\operatorname{Lt}}\left[\frac{x^{4} \sin \left(\frac{1}{x}ight)+x^{2}}{1+|x|^{3}}ight]$\ $=\underset{x ightarrow-\infty}{\operatorname{Lt}} \frac{x^{3}}{1+|x|^{3}}\left[x \sin \left(\frac{1}{x}ight)+\frac{1}{x}ight]$\ $=\underset{x ightarrow-\infty}{\operatorname{Lt}} \frac{x^{3}}{|x|^{3}}\left[\frac{1}{1+\frac{1}{|x|^{3}}}ight]\left[x \sin \left(\frac{1}{x}ight)+\frac{1}{x}ight]$\ Now $\underset{x ightarrow-\infty}{\operatorname{Lt}} \frac{x^{3}}{|x|^{3}}=-1$ and $\underset{x ightarrow-\infty}{\operatorname{Lt}} \frac{1}{|x|^{3}}=0$\ So that $\operatorname{Lt}_{x ightarrow-\infty} \frac{1}{1+\frac{1}{|x|^{3}}}=1$\ Also $\underset{x ightarrow-\infty}{\operatorname{Lt}} x \sin \left(\frac{1}{x}ight)=\underset{y ightarrow 0}{\operatorname{Lt}} \frac{\sin y}{y}=1$\ The required limit $=-1 	imes 1 	imes 1=-1$\