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Question Number 108228 by bemath last updated on 15/Aug/20

$$\frac{⋎\mathcal{B}e\mathcal{M}ath⋎}{⋔}$$$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n+\ln a}{n}\right)}^{\frac{n}{b}}?$$

Answered by Dwaipayan Shikari last updated on 15/Aug/20

$$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{loga}{n})}^{\frac{nloga}{loga}.\frac{1}{b}}$$$$=e^{\frac{loga}{b}}=a^{\frac{1}{b}}$$

Answered by john santu last updated on 15/Aug/20

$$\frac{\divideontimes JS\divideontimes }{\mathrm{♡}}$$$$L=\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n+\ln a}{n}\right)}^{\frac{n}{b}}$$$$\ln L=\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{b}\ln (1+\frac{\ln a}{n})$$$$setx=\frac{1}{n}\to \{\begin{array}{l}n\to \mathrm{\infty }\\ x\to 0\end{array}$$$$\ln L=\underset{x\to 0}{\lim }\frac{\ln (1+x.\ln a)}{bx}$$$$by{L}^{\prime }Hopitalrule$$$$\ln L=\underset{x\to 0}{\lim }\frac{\left(\frac{\ln a}{1+x.\ln a}\right)}{b}=\underset{x\to 0}{\lim }\left(\frac{\ln a}{b(1+x.\ln a)}\right)$$$$\ln L=\frac{\ln a}{b}=\ln {\left(a\right)}^{\frac{1}{b}}$$$$thereforeL=a^{\frac{1}{b}}$$

Answered by bemath last updated on 16/Aug/20

$$\underset{n\to \mathrm{\infty }}{\lim }{\left[{(1+\frac{1}{\left(\frac{n}{\ln a}\right)})}^{\frac{n}{\ln a}}\right]}^{\frac{1}{b}.\frac{\ln a}{1}}$$$$=e{}^{\frac{\ln a}{b}}=e^{\ln {\left(a\right)}^{\frac{1}{b}}}={\left(a\right)}^{\frac{1}{b}}=\sqrt[b]{a}.$$

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