JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

Let

f(x)=\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^2}

for

x \neq \pi / 6

. Then

\lim _{x \rightarrow \pi / 6} f(x)

View answer and explanation

Correct option: B

\frac{1}{36}

Option B (

\frac{1}{36}

) is correct. By using trigonometric identities and L'Hopital's rule, the limit can be evaluated to

\frac{1}{36}

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