Question

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

• A. 1
• B. -1
• C. 0
• D. ∞

Answer 1

The correct option is B

-1

$\underset{x\to \infty }{lim}\left[\frac{x^{4}sin\left(1/x\right)+x^2}{1+|x{|}^3}\right]=\underset{y\to \infty }{lim}\frac{-y^{4}sin\frac{1}{y}+y^2}{1+y^3}[\text{putting x=-y;asx}\to -\infty ,\text{y}\to \infty ].\underset{y\to \infty }{lim}\frac{-\left(\frac{sin\left(1/y\right)}{1/y}\right)+\frac{1}{y}}{1+\frac{1}{y^3}}=\frac{-1+0}{1+0}=-1$