I want to prove that <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJ

Joel French

Joel French

Answered question

2022-07-07

I want to prove that
lim x 1 log ( 1 + 1 x ) 1 1 + x = + .

Answer & Explanation

Melina Richard

Melina Richard

Beginner2022-07-08Added 14 answers

Let
u = 1 x + 1
and then x = u + 1 u . So
log ( 1 + 1 x ) 1 1 + x = u log ( u + 1 )
and hence if x 1 , then u . For u > 6
e u = 1 + u + u 2 2 + u 3 3 ! + > 1 + u + u 2 2 + u 3 3 ! > u 2 + 2 u + 1 = ( u + 1 ) 2 .
So
u log ( u + 1 ) > log ( u + 1 )
and hence
lim x 1 log ( 1 + 1 x ) 1 1 + x = lim u ( u ln ( u + 1 ) ) = .
Lena Bell

Lena Bell

Beginner2022-07-09Added 4 answers

If you set 1 + x = u, then x = u 1 and
1 + 1 x = 1 1 u + 1 = u u + 1
so your limit becomes
lim u 0 + ( log u log ( 1 + u ) + 1 u ) = lim u 0 + ( u log u + 1 u log ( 1 + u ) )
and you can use
lim u 0 + u log u = 0 , lim u 0 log ( 1 + u ) = 0

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