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

Question Number 28084 by tawa tawa last updated on 20/Jan/18

\lim_{x \to \infty} (x - \log_{e} x)

Commented by çhëý böý last updated on 20/Jan/18

\lim_{x \to\propto} (x - l n x)

\lim_{x \to\propto} ({l n e}^{x} - l n x)

\lim_{x \to \infty} l n (\frac{e^{x}}{l n x})

l^{'} h o p i t a l

\lim_{x \to \infty} \frac{e^{x}}{\frac{1}{x}} = \frac{e^{\infty}}{0} = D N E

Commented by tawa tawa last updated on 20/Jan/18

God bless you sir

Commented by abdo imad last updated on 20/Jan/18

w e k n o w t h a t D_{l n x} =] 0, + \infty [s o

{l i m}_{x \to + \infty^{}}^{} (x - l n x) = {l i m}_{x \to + \infty} x (1 - \frac{l n x}{x}) = {l i m}_{x \to + \infty} x

= + \infty b e c a u s e {l i m}_{x \to + \infty^{}} \frac{l n x}{x} = 0 .

Commented by abdo imad last updated on 20/Jan/18

l n (\frac{e^{x}}{l n x}) = x - l n (l n x) \neq l n (e^{x}) - l n x y o u h a v e c o m m i t e d a

e r r o r s i r .

Commented by çhëý böý last updated on 20/Jan/18

y e a h i m a d e m i s t a k e i t s h o u l d b e

l n \frac{{(e^{x})}^{^{'}}}{{(x)}^{^{'}}} = \frac{e^{x}}{1} = e^{\infty} = \infty

t h a n k u f o r d r e c t i f i c a t i o n s i r

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

God bless you sir