solve-without-using-l-hopital-lim-x-e-ln-x-1-e-x-1-

Question Number 91613 by M±th+et+s last updated on 01/May/20

s o l v e w i t h o u t u s i n g l^{'} h o p i t a l

\underset{x \to e}{l i m} \frac{l n (x) - 1}{\frac{e}{x} - 1}

Commented by abdomathmax last updated on 01/May/20

c h a n g e m e n t \frac{e}{x} = t g i v e \frac{x}{e} = \frac{1}{t} \Rightarrow x = \frac{e}{t}

{l i m}_{x \to e} \frac{l n (x) - 1}{\frac{e}{x} - 1} = {l i m}_{t \to 1} \frac{l n (\frac{e}{t}) - 1}{t - 1}

= {l i m}_{t \to 1} \frac{- l n t}{t - 1} = - {l i m}_{t \to 1} \frac{l n (t)}{t - 1} = - 1

Commented by M±th+et+s last updated on 01/May/20

t h a n k y o u s i r

Commented by arcana last updated on 02/May/20

why \lim_{t \to 1} \frac{\ln (t)}{t - 1} = 1 without L^{'} Hopital ?

Commented by mathmax by abdo last updated on 02/May/20

{l i m}_{t \to t_{0}} \frac{f (t) - f (t_{0})}{t - t_{0}} = f^{^{'}} (t_{0}) s o

{l i m}_{t \to 1} \frac{l n (t)}{t - 1} = {l i m}_{t \to 1} \frac{l n (t) - l n (1)}{t - 1} = {l n}^{^{'}} (1) = 1

({l n}^{^{'}} (t) = \frac{1}{t})

Answered by mr W last updated on 02/May/20

l e t t = \frac{e}{x} - 1

x \to e \Rightarrow t \to 0

\Rightarrow x = \frac{e}{1 + t}

\underset{x \to e}{l i m} \frac{l n (x) - 1}{\frac{e}{x} - 1}

= \lim_{t \to 0} \frac{\ln \frac{e}{1 + t} - 1}{t}

= - \lim_{t \to 0} \frac{\ln (1 + t)}{t}

= - \lim_{t \to 0} {\ln {(1 + t)}^{\frac{1}{t}}}

= - \lim_{n \to \infty} {\ln {(1 + \frac{1}{n})}^{n}} \leftarrow d e f i n i t i o n o f e

= - \ln e

= - 1

Commented by M±th+et+s last updated on 02/May/20

g o d b l e s s y o u s i r