MCQ
If $f(x) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $, then $\mathop {\lim }\limits_{x \to \infty } f(x)$is
- A$0$
- B$\infty $
- ✓$1$
- DNone of these
✓
Answer
Correct option: C.
$1$
c
(c) $\mathop {\lim }\limits_{x \to \infty } \,\,f(x) = \,\mathop {\lim }\limits_{x \to \infty } \,\sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $
(c) $\mathop {\lim }\limits_{x \to \infty } \,\,f(x) = \,\mathop {\lim }\limits_{x \to \infty } \,\sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $
$= \mathop {\lim }\limits_{x \to \infty } \,\,\sqrt {\frac{{1 - \frac{{\sin x}}{x}}}{{1 + \frac{{{{\cos }^2}x}}{x}}}} $
$ = \sqrt {\frac{{1 - 0}}{{1 + 0}}} = 1$,
$\left( {\because \,\,\,\frac{{\sin x}}{x} \to 0,\frac{{{{\cos }^2}x}}{x}\, \to 0\,\,{\text{as }}x \to \infty } \right)$.
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