Question

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

• A.
• B.
• C.
• D. does not exist

Answer 1

The correct option is C

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

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

As $x\to -\infty$

$\frac{x^3}{1+|x{|}^3}=-1$

$xsin\left(\frac{1}{x}\right)+\frac{1}{x}=1$

$\Rightarrow \underset{x\to -\infty }{lim}\frac{x^3}{1+|x{|}^3}(xsin\frac{1}{x}+\frac{1}{x})=-1$