Find-1-3n-1-n-Limit-as-n-Please-help-

Question Number 5616 by sanusihammed last updated on 22/May/16

F i n d

{[1 + 3 n^{- 1}]}^{n}

L i m i t a s n \to \infty

P l e a s e h e l p .

Answered by FilupSmith last updated on 22/May/16

L = \lim_{n \to \infty} {(1 + \frac{3}{n})}^{n}

L = \lim_{n \to \infty} \exp (n \ln (1 + \frac{3}{n}))

L = \lim_{n \to \infty} \exp (\frac{\ln (1 + \frac{3}{n})}{\frac{1}{n}})

L^{'} {H o p i t a l}^{'} s L a w

L = \lim_{n \to \infty} \exp (\frac{[\frac{- 6 n^{- 2}}{1 + 3 n^{- 1}}]}{- 2 n^{- 2}})

L = \lim_{n \to \infty} \exp (\frac{6}{n^{2}} \times \frac{1}{1 + \frac{3}{n}} \times \frac{n^{2}}{2})

L = \lim_{n \to \infty} \exp (3 \times \frac{1}{1 + \frac{3}{n}})

L = \lim_{n \to \infty} \exp (3 \times \frac{1}{\frac{n + 3}{n}})

L = \lim_{n \to \infty} \exp (3 \times \frac{n}{n + 3})

L^{'} {H o p i t a l}^{'} s L a w

L = \lim_{n \to \infty} \exp (3 \times \frac{1}{1})

L = e^{3}

∴ \lim_{n \to \infty} {(1 + \frac{3}{n})}^{n} = e^{3}

Commented by sanusihammed last updated on 22/May/16

T h a n k s . i a p p r e c i a t e