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

The correct option is B $0$

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