JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

\lim _{x \rightarrow \infty} \sqrt{\frac{x+\sin x}{x-\cos x}}

is equal to

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 x}}{{x}}

approach 0, leaving

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

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\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

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\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}