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 $
  • $1$
  • D
    None 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}}} $

$= \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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