Question

${lim}_{x\to \infty }\frac{e^x({\left(2^{x^n}\right)}^{\frac{1}{e^x}}-{\left(3^{x^n}\right)}^{\frac{1}{e^x}})}{x^n},nϵN$ is equal to

• A. $0$
• B. $\ln \left(\frac{2}{3}\right)$
• C. $\ln \left(\frac{3}{2}\right)$
• D. $Noneofthese$

Answer 1

Answer: (B)

$\ln \left(\frac{2}{3}\right)$

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

Apply L-Hospital Rule

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