Question

The value of $\underset{x\to 0}{\mathrm{Lim}}\frac{\ln \left(\sec (ex)\sec \left(e^{2}x\right)\dots \dots .\sec \left(e^{50}x\right)\right)}{e^2-e^{2\cos x}}$ is equal to

• A. $\frac{e^{100}-1}{\left(e^2-1\right)}$
• B. $\frac{e^{100}-1}{2\left(e^2-1\right)}$
• C. $\frac{2\left(e^{50}-1\right)}{e^2-1}$
• D. $\frac{e^2\left(e^{100}-1\right)}{2\left(e^2-1\right)}$

Answer 1

Answer: (B)

$\underset{x\to 0}{\mathrm{Lim}}\frac{\ln \left(\sec (ex)\sec \left(e^{2}x\right)\cdot \sec \left(e^{3}x\right)\dots \dots \cdot \sec \left(e^{50}x\right)\right)}{e^{2\cos x}\frac{\left(e^{2-2\cos x}-1\right)}{2-2\cos x}\cdot \frac{2(1-\cos x)}{x^2}\cdot x^2}$
$=\underset{x\to 0}{\mathrm{Lim}}\frac{\ln \left(\sec (ex)\sec \left(e^{2}x\right)\dots \dots \sec \left(e^{50}x\right)\right)}{e^2{x}^2}$
$=\underset{x\to 0}{\mathrm{Lim}}\frac{\ln \left(1+\sec (ex)\sec \left(e^{2}x\right)\dots \dots \sec \left(e^{50}x\right)-1\right)}{e^2{x}^2}$
$=\underset{x\to 0}{\mathrm{Lim}}\frac{\sec (ex)\sec \left(e^{2}x\right)\cdot \sec \left(e^{3}x\right)\dots \dots \cdot \sec \left(e^{50}x\right)-1}{e^2{x}^2}$
Apply L'hospital Rule, get
$L=\frac{e^2+e^4+e^6+\dots \dots +e^{100}}{2e^2}=\frac{e^2\left({\left(e^2\right)}^{50}-1\right)}{2e^2\left(e^2-1\right)}=\frac{e^{100}-1}{2\left(e^2-1\right)}$