prove-that-lim-x-0-ln-x-ln-1-x-lim-x-1-ln-x-ln-1-x-

Question Number 34160 by candre last updated on 01/May/18

p r o v e t h a t

\underset{x \to 0^{+}}{\lim l n} x \cdot \ln (1 + x) = \underset{x \to 1}{\lim l n} x \cdot \ln (1 + x)

Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

s e e t h e g r a p h L H S l i m i t i s z e r o a l s o R H S l i m i t

a l s o z e r o

Commented by abdo mathsup 649 cc last updated on 02/May/18

w e h a v e {l i m}_{x \to 0^{+}} l n (x) l n (1 + x)

= {l i m}_{x \to 0^{+}} (x l n x) \frac{l n (1 + x)}{x} = 0 b e c a u s e {l i m}_{x \to 0^{+}} x l n (x) = 0

{l i m}_{x \to 0^{+}} \frac{l n (1 + x)}{x} = 1 f r o m a n o t h e r s i d e

{l i m}_{x \to 1} l n (x) . l n (1 + x) = l n (1) l n (2) = 0 b e c a u s e

t h e f u n c t i o n l n (x) i s c o n t i n u e a t 1 a n d 2 .

Answered by MJS last updated on 02/May/18

right side :

\ln 1 = 0

0 \ln 2 = 0

left side :

e^{\ln x \ln (x + 1)} = {(x + 1)}^{\ln x}

\lim_{x \to 0^{+}} {(x + 1)}^{\ln x} = 1^{- \infty} = 1 \Rightarrow

\Rightarrow \underset{x \to 0^{+}}{\lim l n} x \ln (x + 1) = 0

Facebook Tweet Pin