Question

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

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

Answer 1

Answer: (B)

$0$

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

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

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

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

Hence, option 'B' is correct.