Question

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

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

Answer 1

Answer: (B)

$L=\underset{x\to \infty }{\lim }\frac{{\left(2^{x^n}\right)}^{\frac{1}{e^x}}-{\left(3^{x^n}\right)}^{\frac{1}{e^x}}}{x^n}$
$=\underset{x\to \infty }{\lim }\frac{(3{)}^{\frac{x^n}{e^x}}\left({\left(\frac{2}{3}\right)}^{\frac{x^n}{e^x}}-1\right)}{x^n}$
Now, $\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=\underset{x\to \infty }{\lim }\frac{n!}{e^x}=0$
(Differentiating numerator and denominator $n$ times for L'Hospital's rule)
Hence, $L=\underset{x\to \infty }{\lim }(3{)}^{\frac{x^n}{x^x}}\underset{x\to \infty }{\lim }\frac{\left({\left(\frac{2}{3}\right)}^{\frac{x^n}{e^x}}-1\right)}{\frac{x^n}{e^x}}\underset{x\to \infty }{\lim }\frac{1}{e^x}$
$=1\times \log (2/3)\times 0=0$