MCQ
$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}} = . . .$
- ✓$0$
- B$1$
- C$-1$
- Dએકપણ નહી.
$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\frac{{\sin t}}{t} - 1}}{{t - 1}} = \frac{{1 - 1}}{{0 - 1}} = 0,$
which is given in $(a)$.
Aliter : $\mathop {\lim }\limits_{x \to \infty } \,\,\frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - \,\,|x|}}$
$ = \mathop {\lim }\limits_{x \to \infty } \,\,\frac{{{x^2}\,\left( {\frac{1}{x} - \frac{1}{{3\,\,!}}\frac{1}{{{x^3}}} + ....} \right) - x}}{{1 - |x|}}$,
$ = \mathop {\lim }\limits_{x \to \infty } \,\,\frac{{\left( {x - \frac{1}{{6x}} + .... - x} \right)}}{{1 - |x|}}$
$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{{\frac{1}{{6x}} - {\rm{terms \,containing \,powers\, of\, }}\frac{1}{x}}}{{|x| - 1}} = 0.$
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