Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $\infty$

Answer 1

Answer: (B)

$-1$

$\underset{x\to -\infty }{lim}\left[\frac{x^4\sin \left(\frac{1}{x}\right)+x^2}{1+{\left|x\right|}^3}\right]=\underset{y\to \infty }{lim}\frac{-y^4\sin \frac{1}{y}+y^2}{1+y^3}$

$[$ Substitute $x=-y;$ as $x\to -\infty ,y\to \infty ]$

$=\underset{y\to \infty }{lim}\frac{-\left(\frac{\sin \frac{1}{y}}{\frac{1}{y}}\right)+\frac{1}{y}}{1+\frac{1}{y^3}}$..................................$\left(dividingnumeratoranddenominatorby{y}^{3}andrearranging\right)$

$=\frac{-1+0}{1+0}=-1$