Question

The value of $\underset{x\to -\infty }{\lim }\left\{\frac{x^4\sin \left(\frac{1}{x}\right)+x^2}{1+\left|x^3\right|}\right\}$ is

• A. 1
• B. -1
• C. 0
• D. $\infty$

Answer 1

Answer: (B)

We have, $\underset{x\to -\infty }{\lim }\left\{\frac{x^4\sin \left(\frac{1}{x}\right)+x^2}{1+\left|x^3\right|}\right\}$
$=\underset{y\to \infty }{\lim }\frac{y^4\sin \left(\frac{-1}{y}\right)+y^2}{1+\left|-y^3\right|}$,
where $y=-x$
$=\underset{y\to \infty }{\lim }\frac{-y^4\sin \left(\frac{1}{y}\right)+y^2}{1+y^3}$
$=\underset{y\to \infty }{\lim }\frac{-y\sin \left(\frac{1}{y}\right)+\frac{1}{y}}{\frac{1}{y^3}+1}$
$=\frac{-1+0}{0+1}=-1$