Question

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

• A. ${\log }_n\left(\frac{2}{3}\right)$
• B. $0$
• C. $n{\log }_n\left(\frac{2}{3}\right)$
• D. It does not exist

Answer 1

The correct option is B $0$

Let $L=\underset{x\to \infty }{lim}\frac{{\left(3\right)}^{\frac{x^n}{e^x}}({\left(\frac{2}{3}\right)}^{\frac{x^n}{e^x}}-1)}{x^n}$

Now,
$\underset{x\to \infty }{lim}\frac{x^n}{e^x}=\underset{x\to \infty }{lim}\frac{n!}{e^x}=0\left(\text{Differentiating numerator and denominator n times for L' Hospital's rule}\right)$

Hence, $L=\underset{x\to \infty }{lim}{\left(3\right)}^{\frac{x^n}{e^x}}.\underset{x\to \infty }{lim}\frac{{\left(\frac{2}{3}\right)}^{\frac{x^n}{e^x}}-1}{\frac{x^n}{e^x}}.\underset{x\to \infty }{lim}\frac{1}{e^x}$ $=1\times \text{log}\left(\frac{2}{3}\right)\times 0=0$