Question

The value of $\underset{x\to \infty }{\mathrm{Limit}}\frac{{\left(2^{x^n}\right)}^{\frac{1}{e^x}}-{\left(3^{x^n}\right)}^{\frac{1}{e^x}}}{x^n}($ 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)

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