is
A) 0
B) 1
C) -1
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{2}}\sin \frac{1}{x}-x}{1-\left| x \right|}\] \[=\,\,\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{2}}\left( \frac{1}{x}-\frac{1}{3!}\frac{1}{{{x}^{3}}}+... \right)-x}{1-\left| x \right|},\,\,\left[ as\,\,x\to \infty ;\,\frac{1}{x}\to 0 \right]\] \[=\,\,\underset{x\to \infty }{\mathop{\lim }}\,\,\,\frac{\left( x-\frac{1}{6x}+...-x \right)}{1-\left| x \right|}\] \[=\,\,\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{\frac{1}{6x}-terms\,\,containing\,\,power\,\,of\frac{1}{x}}{\left| x \right|-1}=0\]
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