lim-n-1-c-1-1-n-n-with-0-c-1-gt-0-

Question Number 54566 by mr W last updated on 06/Feb/19

\lim_{n \to \infty} {(\frac{1 + c}{1 + μ^{\frac{1}{n}}})}^{n} = ?

w i t h 0 ⩽ c ⩽ 1, μ > 0

Commented by maxmathsup by imad last updated on 06/Feb/19

l e t A_{n} = {(\frac{1 + e}{1 + μ^{\frac{1}{n}}})}^{n} \Rightarrow A_{n} = {(\frac{1 + e + μ^{\frac{1}{n}} - μ^{\frac{1}{n}}}{1 + μ^{\frac{1}{n}}})}^{n} = {(1 + \frac{e - μ^{\frac{1}{n}}}{1 + μ^{\frac{1}{n}}})}^{n}

= e^{n l n (1 + \frac{e - μ^{\frac{1}{n}}}{1 + μ^{\frac{1}{n}}}) b u t μ^{\frac{1}{n}} = e^{\frac{l n (μ)}{n}} \sim 1 + \frac{l n (μ)}{n}} \Rightarrow e - μ^{\frac{1}{n}} \sim e - 1 - \frac{l n (μ)}{n} a n d

1 + μ^{\frac{1}{n}} = 2 + \frac{l n (μ)}{n} \Rightarrow 1 + \frac{e - μ^{\frac{1}{n}}}{1 + μ^{\frac{1}{n}}} \sim 1 + \frac{e - 1}{2} = \frac{1 + e}{2} \Rightarrow A_{n} \sim e^{n l n (\frac{1 + e}{2})} \to + \infty (n \to + \infty)

(l o o k t h a t l n (\frac{1 + e}{2}) > 0) w e c o n s i d e r μ a n d e f i x e d . .

Commented by mr W last updated on 06/Feb/19

t h a n k y o u s i r!

c ⩽ 1

t h e r e f o r e \ln (\frac{1 + c}{2}) ⩽ 0

i t s e e m s i f c < 1, {(\frac{1 + e}{1 + μ^{\frac{1}{n}}})}^{n} \to 0

w h a t i f c = 1 ?, i . e . {(\frac{2}{1 + μ^{\frac{1}{n}}})}^{n} \to ?

Commented by maxmathsup by imad last updated on 06/Feb/19

y o u a r e r i g h t s i r i h a v e c o m m i t e d a e r r o r o f c a l c u l u s w e h a v e

0 < e < 1 \Rightarrow 1 < 1 + e < 2 \Rightarrow \frac{1}{2} < \frac{1 + e}{2} < 1 \Rightarrow l n (\frac{1 + e}{2}) < 0 \Rightarrow {l i m}_{n \to + \infty} A_{n} = 0

f o r g i v e m e i w a s t i r e d \dots