The value of $\lim _{x \rightarrow 0} \frac{\log _{e}\left(\sec (e x) \cdot \sec \left(e^{2} x\right) \cdot \ldots \cdot \sec \left(e^{10} x\right)\right)}{e^{2}-e^{2 \cos x}}$ is equal to

(C) $\frac{\left(\mathrm{e}^{20}-1\right)}{2\left(\mathrm{e}^{2}-1\right)}$

(A) $\frac{\left(\mathrm{e}^{10}-1\right)}{2 \mathrm{e}^{2}\left(\mathrm{e}^{2}-1\right)}$

(B) $\frac{\left(\mathrm{e}^{20}-1\right)}{2 \mathrm{e}^{2}\left(\mathrm{e}^{2}-1\right)}$

(D) $\frac{\left(\mathrm{e}^{10}-1\right)}{2\left(\mathrm{e}^{2}-1\right)}$

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Limits

The value of

lim_{x \to 0} \frac{l o g _{e} ( s e c ( e x ) \cdot s e c ( e ^{2} x ) \cdot \dots \cdot s e c ( e ^{10} x ) )}{e ^{2} - e ^{2 c o s x}}

is equal to

Answer: (C)

Each factor gives

ln sec (e^{k} x) = - ln cos (e^{k} x) \approx \frac{1}{2} (e^{k} x)^{2}

x \to 0

, so the numerator behaves like

\frac{x ^{2}}{2} \sum_{k = 1}^{10} e^{2 k}

. For the denominator,

e^{2} - e^{2 c o s x} = e^{2} (1 - e^{2 (c o s x - 1)})

, and since

cos x - 1 \approx - \frac{x ^{2}}{2}

1 - e^{2 (c o s x - 1)} \approx x^{2}

, so the denominator behaves like

e^{2} x^{2}

. Hence the limit equals

\frac{1}{2 e ^{2}} \sum_{k = 1}^{10} e^{2 k} = \frac{1}{2 e ^{2}} \cdot \frac{e ^{2} ( e ^{20} - 1 )}{e ^{2} - 1} = \frac{e ^{20} - 1}{2 ( e ^{2} - 1 )}

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option