Question

The value of $\underset{x \rightarrow 0}{\text{Lim}} \frac{\ln \left(\sec (e x) \sec \left(e^2 x\right) \ldots \ldots . \sec \left(e^{50} x\right)\right)}{e^2-e^{2 \cos x}}$ 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 \rightarrow 0}{\text{Lim}} \frac{\ln \left(\sec (e x) \sec \left(e^2 x\right) \cdot \sec \left(e^3 x\right) \ldots \ldots \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 \rightarrow 0}{\text{Lim}} \frac{\ln \left(\sec (e x) \sec \left(e^2 x\right) \ldots \ldots \sec \left(e^{50} x\right)\right)}{e^2 x^2} $
$=\underset{x \rightarrow 0}{\text{Lim}} \frac{\ln \left(1+\sec (e x) \sec \left(e^2 x\right) \ldots \ldots \sec \left(e^{50} x\right)-1\right)}{e^2 x^2}$
$=\underset{x \rightarrow 0}{\text{Lim}} \frac{\sec (e x) \sec \left(e^2 x\right) \cdot \sec \left(e^3 x\right) \ldots \ldots \cdot \sec \left(e^{50} x\right)-1}{e^2 x^2}$
Apply L'hospital Rule, get
$L =\frac{ e ^2+ e ^4+ e ^6+\ldots \ldots+ e ^{100}}{2 e ^2}=\frac{ e ^2\left(\left( e ^2\right)^{50}-1\right)}{2 e ^2\left( e ^2-1\right)}=\frac{ e ^{100}-1}{2\left( e ^2-1\right)}$