BeMath-lim-n-n-ln-a-n-n-b-

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

\frac{⋎ B e M a t h ⋎}{⋔}

\lim_{n \to \infty} {(\frac{n + \ln a}{n})}^{\frac{n}{b}} ?

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

\lim_{n \to \infty} {(1 + \frac{l o g a}{n})}^{\frac{n l o g a}{l o g a} . \frac{1}{b}}

= e^{\frac{l o g a}{b}} = a^{\frac{1}{b}}

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

\frac{⋇ J S ⋇}{♡}

L = \lim_{n \to \infty} {(\frac{n + \ln a}{n})}^{\frac{n}{b}}

\ln L = \lim_{n \to \infty} \frac{n}{b} \ln (1 + \frac{\ln a}{n})

s e t x = \frac{1}{n} \to {\begin{cases} n \to \infty \\ x \to 0 \end{cases}

\ln L = \lim_{x \to 0} \frac{\ln (1 + x . \ln a)}{b x}

b y L^{'} H o p i t a l r u l e

\ln L = \lim_{x \to 0} \frac{(\frac{\ln a}{1 + x . \ln a})}{b} = \lim_{x \to 0} (\frac{\ln a}{b (1 + x . \ln a)})

\ln L = \frac{\ln a}{b} = \ln {(a)}^{\frac{1}{b}}

t h e r e f o r e L = a^{\frac{1}{b}}

Answered by bemath last updated on 16/Aug/20

\lim_{n \to \infty} {[{(1 + \frac{1}{(\frac{n}{\ln a})})}^{\frac{n}{\ln a}}]}^{\frac{1}{b} . \frac{\ln a}{1}}

= e^{\frac{\ln a}{b}} = e^{\ln {(a)}^{\frac{1}{b}}} = {(a)}^{\frac{1}{b}} = \sqrt[b]{a} .

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