fond-lim-n-n-a-ln-1-e-n-e-n-with-a-gt-0-

Question Number 35833 by abdo mathsup 649 cc last updated on 24/May/18

f o n d {l i m}_{n \to + \infty} n^{a} {l n (1 + e^{- n}) - e^{- n}} w i t h a > 0

Commented by prof Abdo imad last updated on 25/May/18

w e h a v e f o r ∣ x ∣< 1 {l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}

\Rightarrow l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} x^{n}}{n}

s o l n (1 + x) = x - \frac{x^{2}}{2} + o (x^{3}) (x \to 0) \Rightarrow

l n (1 + e^{- n}) = e^{- n} - \frac{e^{- 2 n}}{2} + o (e^{- 3 n}) (n \to + \infty)

n^{a} {l n (1 + e^{- n}) - e^{- n}} \sim - \frac{1}{2} n^{a} e^{- 2 n} (n \to + \infty)

b u t {l i m}_{n \to + \infty} n^{a} e^{- 2 n} = {l i m}_{n \to + \infty} e^{a l n (n) - 2 n}

= {l i m}_{n \to + \infty} e^{n {a \frac{l n (n)}{n} - 2}} = {l i m}_{n \to + \infty} e^{- 2 n} = 0 s o

f o r a l l a > 0 {l i m}_{n \to + \infty} n^{a} {l n (1 + e^{- n}) - e^{- n}} = 0

Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18

w h e n n \to \infty t h e n e^{- n} \to 0

s o g i v e n l i m i t (\infty \times 0) f o r m

= \frac{l i m}{n \to \infty} \frac{{l n (1 + e^{- n}) - e^{- n}}}{\frac{1}{n^{a}}} (\frac{0}{0}) f o r m

u s i n g l h o s p i t s l

= \frac{l i m}{n \to \infty} \times \frac{{\frac{1}{(1 + e^{- n})} \times e^{- n} \times - 1} - {e^{- n} \times (- 1)}}{- a \times n^{- a - 1}}

= \frac{l i m}{n \to \infty} \times \frac{{\frac{- e^{- n}}{1 + e^{- n}}} + e^{- n}}{- a \times n^{- a - 1}}

= \frac{l i m}{n \to \infty} \times \frac{{- e^{- n} + e^{- n} + e^{- 2 n}}{(1 + e^{- n}) \times (- a \times n^{- a - 1})}

= (\frac{- 1}{a}) \times \frac{l i m}{n \to \infty} \frac{e^{- 2 n}}{1 + e^{- n}} \times n^{a + 1}

= (\frac{- 1}{a}) \times \frac{l i m}{n \to \infty} \frac{1}{e^{2 n} + e^{n}} \times n^{a + 1}

= (\frac{- 1}{a}) \times \frac{l i m}{n \to \infty} \frac{1}{e^{n} + 1} \times \frac{n^{a + 1}}{e^{n}}

= (\frac{-}{a}) \times \frac{l i m}{n \to \infty} \times \frac{\frac{1}{e^{2 n} + e^{n}}}{\frac{1}{n^{a + 1}}}

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