JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim \limits_{{x \rightarrow \pi / 6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}

equals

View answer and explanation

Correct option: B

\frac{1}{\sqrt{3}}

Option B (

\frac{1}{\sqrt{3}}

) is correct because the limit can be evaluated using L'Hopital's rule or by rewriting the expression using trigonometric identities.

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\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

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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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