JEE Main 202628 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002The value of limx→0loge(sec(ex)⋅sec(e2x)⋅…⋅sec(e10x))e2−e2cosx\lim_{x\to0}\frac{\log_e\left(\sec(ex)\cdot\sec(e^2x)\cdot\ldots\cdot\sec(e^{10}x)\right)}{e^2-e^{2\cos x}}limx→0e2−e2cosxloge(sec(ex)⋅sec(e2x)⋅…⋅sec(e10x)) is equal toAOption A. e20−12e2(e2−1)\frac{e^{20}-1}{2e^2(e^2-1)}2e2(e2−1)e20−1BOption B. e20−12(e2−1)\frac{e^{20}-1}{2(e^2-1)}2(e2−1)e20−1COption C. e10−12(e2−1)\frac{e^{10}-1}{2(e^2-1)}2(e2−1)e10−1DOption D. e10−12e2(e2−1)\frac{e^{10}-1}{2e^2(e^2-1)}2e2(e2−1)e10−1CheckSkip · show solution
