Question-52208

Question Number 52208 by SUJIT420 last updated on 04/Jan/19

Commented by maxmathsup by imad last updated on 04/Jan/19

i t s n o t c i t s e .

Commented by maxmathsup by imad last updated on 04/Jan/19

l e t U (x) = {(1 + x)}^{\frac{1}{x}} - e a n d V (x) = x w e h a v e

U (x) = e^{\frac{1}{x} l n (1 + x)} - e \Rightarrow {l i m}_{x \to 0} U (x) = 0 h o s p i t a l t h e o r e m g i v e

{l i m}_{x \to 0} \frac{U (x)}{x} = {l i m}_{x \to 0} U^{^{'}} (x) b u t U^{^{'}} (x) = {(\frac{l n (1 + x)}{x})}^{^{'}} U (x)

= \frac{\frac{x}{1 + x} - l n (1 + x)}{x^{2}} U (x) = \frac{x - (1 + x) l n (1 + x)}{x^{2}} U (x) b u t

l n (1 + x) \sim x \Rightarrow x - (1 + x) l n (1 + x) \sim x - x (1 + x) = - x^{2} \Rightarrow U^{^{'}} (x) \sim - U (x) (x \to 0)

w e h a v e {l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = 1 - x + o (x^{2}) \Rightarrow l n (1 + x) = x - \frac{x^{2}}{2} + o (x^{3}) \Rightarrow

\frac{l n (1 + x)}{x} = 1 - \frac{x}{2} + o (x^{2}) \Rightarrow e^{\frac{l n (1 + x)}{x}} = e^{1 - \frac{x}{2} + o (x^{2})} = e (1 - \frac{x}{2} + o (x^{2})) \Rightarrow

e^{\frac{l n (1 + x)}{x}} - e = - \frac{e x}{2} + o (x^{2}) \Rightarrow \frac{U (x)}{x} = - \frac{e}{2} + o (x) \Rightarrow {l i m}_{x \to 0} \frac{U (x)}{x} = - \frac{e}{2}

a n d w e s e e t h a t h o s p i t a l t h e o r e m d o n t w o r k g o o d i n t h i s c a s e .

Commented by afachri last updated on 05/Jan/19

is it possible to transform {(1 + x)}^{\frac{1}{x}} - e into Taylor

serie Sir ? ? ? because i got stuck to find the

derrivative of {(1 + x)}^{\frac{1}{x}}

Answered by Smail last updated on 05/Jan/19

\underset{x \to 0}{l i m} \frac{{(1 + x)}^{1 / x} - e}{x} = \underset{x \to 0}{l i m} \frac{e^{\frac{l n (1 + x)}{x}} - e}{x}

l n (1 + x) \approx x - \frac{x^{2}}{2} n e a r 0

s o \underset{x \to 0}{l i m} \frac{{(1 + x)}^{1 / x} - e}{x} = \underset{x \to 0}{l i m} \frac{e^{\frac{x - \frac{x^{2}}{2}}{x}} - e}{x}

= \underset{x \to 0}{l i m} \frac{e^{1 - \frac{x}{2}} - e}{x} = \underset{x \to 0}{l i m} \frac{e (e^{- \frac{x}{2}} - 1)}{x}

e^{t} \approx 1 + t n e a r 0

\underset{x \to 0}{l i m} \frac{{(1 + x)}^{1 / x} - e}{x} = \underset{x \to 0}{l i m} \frac{e (1 - \frac{x}{2} - 1)}{x} = \underset{x \to 0}{l i m} \frac{- x e}{2 x}

= \frac{- e}{2}

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