lim-x-1-1-log-e-x-x-x-1-

Question Number 28151 by tawa tawa last updated on 21/Jan/18

\lim_{x \to 1} (\frac{1}{\log_{e} x} - \frac{x}{x - 1})

Commented by çhëý böý last updated on 21/Jan/18

\lim_{x \to 1} (\frac{(x - 1) - x l n x}{(x - 1) l n x}) = \frac{0}{0}

\lim_{x \to 1} (\frac{x - (\frac{x}{x} + l n x)}{l n x + \frac{(x - 1)}{x}})

\lim_{x \to 1} (\frac{x (x - 1 + 1 n x)}{x l n x + (x - 1)}) = \frac{0}{0}

\lim_{x \to 1} (\frac{(x - 1 + l n x) + x (1 - 0 + \frac{1}{x})}{1 + l n x + 1})

\lim_{x \to 1} (\frac{0 + 0 + 2}{1 + 0 + 1}) = \frac{2}{2} = 1

Commented by tawa tawa last updated on 21/Jan/18

God bless you sir

Commented by abdo imad last updated on 21/Jan/18

= {l i m}_{x \to 1} (\frac{1}{l n x} - \frac{x}{x - 1}) l e t u s e t h e c h . x - 1 = t

x \to 1 \Leftrightarrow t \to o l i m (\dots) = {l i m}_{t \to 0} (\frac{1}{l n (1 + t)} - \frac{1 + t}{t})

= {l i m}_{t \to 0} \frac{t - (1 + t) l n (1 + t)}{t l n (1 + t)} l e t t a k e f (t) = t - (1 + t) l n (1 + t)

a n d g (t) = t l n (1 + t) w e h a v e

f^{^{'}} (t) = 1 - (l n (1 + t) + 1) = - l n (1 + t) a n d f^{^{″}} (t) = \frac{- 1}{1 + t}

g^{^{'}} (t) = l n (1 + t) + \frac{t}{1 + t} = l n (1 + t) + 1 - \frac{1}{1 + t} a n d

g^{^{″}} (t) = \frac{1}{1 + t} + \frac{1}{{(1 + t)}^{2}} s o

{l i m}_{x \to 1} (\frac{1}{l n x} - \frac{x}{x - 1}) = {l i m}_{t \to 0} \frac{f^{^{″}} (t)}{g^{^{″}} (t)} = \frac{- 1}{2} .

Commented by abdo imad last updated on 21/Jan/18

w e h a v e u s e d h o s p i t a l t h e o r e m .