JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Analyzing +4 | -1

For

n \in N

, let

f_n(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x)

(1+\sec 4 x) \ldots\left(1+\sec 2^n x\right)

. Then,

\lim \limits_{{x \rightarrow 0}} \frac{f_n(x)}{2 x}

View answer and explanation

Correct option: C

2^{n-1}

Option C,

2^{n-1}

, is correct because by repeatedly applying the identity

1 + \sec x = \frac{1 + \cos x}{\cos x} = \frac{2 \cos^2(x/2)}{\cos x}

, the function

f_n(x)

simplifies to

\tan(2^n x)

, and the limit can be evaluated using L'Hopital's rule or the standard limit

\lim_{x \to 0} \frac{\tan x}{x} = 1

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