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)

0

Answer 2

$\lim_{x\to\infty} \left[\frac{x^{2} \sin\left(\frac{1}{x}\right) -x}{1-\left|x\right|}\right]$ = $\lim_{x\to\infty}\left[\frac{x^{2}\sin\left(\frac{1}{x}\right) -x}{1-x}\right] = \lim_{x\to\infty} \frac{\frac{\sin\left(x^{-1}\right)-1}{x^{-1}}}{x^{-1} -1}$ = $\lim_{y\to0 } \frac{\frac{\sin\,y}{y} -1}{y-1} = \frac{1-1}{0-1}=0$