If $f\left( x \right) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $, then $\mathop {\lim }\limits_{x \to \infty } f\left( x \right)$ is

JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

f\left( x \right) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}}

, then

\lim\limits_{x \to \infty } f\left( x \right)

View answer and explanation

Correct option: C

Option C (1) is correct because as x approaches infinity, the terms

\frac{\sin x}{x}

and

\frac{\cos^2 x}{x}

approach 0, leaving

\sqrt{\frac{1}{1}} = 1

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