Question

The value of $\underset{x \rightarrow \infty} {\text{Limit}} \frac{\left(2^{x^n}\right)^{\frac{1}{e^x}}-\left(3^{x^n}\right)^{\frac{1}{e^x}}}{x^n} \quad($ where $n \in N)$ is

• A. $\ln \left(\frac{2}{3}\right)$
• B. 0
• C. $n \ln \left(\frac{2}{3}\right)$
• D. not defined

Answer 1

Answer: (B)

$l=\underset { x \rightarrow \infty} {\text{Limit}}\frac{2^{\frac{ x ^{ n }}{ e ^{ x }}}-3^{\frac{ x ^{ n }}{ e ^{ x }}}}{ x ^{ n }} $ but
$\underset { x \rightarrow \infty} {\text{Limit}} \frac{ x ^{ n }}{ e ^{ x }}=0 \Rightarrow l=0$ (using L'Hospital's Rule)