A) \[0\]
B) 1
C) ?1
D) None of these
Correct Answer: A
Solution :
Putting \[x=\frac{1}{t},\] the given limit \[=\underset{t\to 0}{\mathop{\lim }}\,\,\,\frac{\frac{\sin t}{t}-1}{t-1}=\frac{1-1}{0-1}=0,\] which is given in . Aliter : \[\underset{x\to \infty }{\mathop{\lim }}\,\,\,\frac{{{x}^{2}}\sin \frac{1}{x}-x}{1-\,\,|x|}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\,\,\frac{{{x}^{2}}\,\left( \frac{1}{x}-\frac{1}{3\,\,!}\frac{1}{{{x}^{3}}}+.... \right)-x}{1-|x|}\], \[\left[ \because \,\,\frac{1}{x}\to 0 \right]\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\,\,\frac{\left( x-\frac{1}{6x}+....-x \right)}{1-|x|}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{\frac{1}{6x}-\text{terms containing powers of }\frac{1}{x}}{|x|-1}=0.\]
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