The value of lim x 0 log e ( sec ( e x ) · sec ( e 2 x ) ? sec ( e 10 x ) ) e 2 - e 2 cos x is equal to [2026]

Question

The value of $\lim_{x \to 0} \frac{\log_{e} ((\sec (e x) \cdot \sec ((e^{2} x)) \dots \sec ((e^{10} x))))}{e^{2} - e^{2 \cos x}}$ is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$\Rightarrow \lim_{x \to 0} \frac{\ln (\sec (e x)) + \ln (\sec (e^{2} x)) + \dots + l n (\sec (e^{10} x))}{e^{2 \cos x} (\frac{e^{2 - 2 \cos x} - 1}{2 - 2 \cos x}) (\frac{2 - 2 \cos x}{x^{2}}) x^{2}}$

$\Rightarrow \lim_{x \to 0} \frac{l n (\sec (e x)) + l n (\sec (e^{2} x)) + \dots + l n (\sec (e^{10} x))}{e^{2} x^{2}}$

$Using L'H rule$

$\Rightarrow \lim_{x \to 0} \frac{e \tan (e x) + e^{2} \tan (e^{2} x) + \dots + e^{10} \tan (e^{10} x)}{2 e^{2} x}$

$= \frac{1}{2 e^{2}} [e^{2} + e^{4} + e^{6} + \dots + e^{20}]$

$= \frac{1}{2} \cdot \frac{e^{2} ({(e^{2})}^{10} - 1)}{e^{2} (e^{2} - 1)}$

$= \frac{1}{2} \cdot \frac{(e^{20} - 1)}{(e^{2} - 1)}$

Solution

Q.
The value of $\lim_{x \to 0} \frac{\log_{e} ((\sec (e x) \cdot \sec ((e^{2} x)) \dots \sec ((e^{10} x))))}{e^{2} - e^{2 \cos x}}$ is equal to [2026]

1 $\frac{(e^{10} - 1)}{2 e^{2} (e^{2} - 1)}$

2 $\frac{(e^{20} - 1)}{2 e^{2} (e^{2} - 1)}$

3 $\frac{(e^{10} - 1)}{2 (e^{2} - 1)}$

4 $\frac{(e^{20} - 1)}{2 (e^{2} - 1)}$

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Solution

Q. The value of limx→0loge(sec(ex)·sec (e2x)⋯sec (e10x))e2-e2cosx is equal to [2026]

1 (e10-1)2e2(e2-1)

2 (e20-1)2e2(e2-1)

3 (e10-1)2(e2-1)

4 (e20-1)2(e2-1)