Question

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

• A. 0
• B. 1
• C. 2
• D. none of these

Answer 1

Answer: (A)

${\lim }_{x\to \infty }\left[\frac{x^2\sin \left(\frac{1}{x}\right)-x}{1-|x|}\right]$
= ${\lim }_{x\to \infty }\left[\frac{x^2\sin \left(\frac{1}{x}\right)-x}{1-x}\right]={\lim }_{x\to \infty }\frac{\frac{\sin \left(x^{-1}\right)-1}{x^{-1}}}{x^{-1}-1}$
= ${\lim }_{y\to 0}\frac{\frac{\sin y}{y}-1}{y-1}=\frac{1-1}{0-1}=0$