Limit of logarithms exponential lim_(x->oo)((ln(x-2))/(ln(x-1)))^(x\ln x). L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?

reevelingw97

reevelingw97

Answered question

2022-11-21

lim x ( ln ( x 2 ) ln ( x 1 ) ) x ln x .
L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?

Answer & Explanation

grizintimbp

grizintimbp

Beginner2022-11-22Added 16 answers

Let
f ( x ) = ( ln ( x 2 ) ln ( x 1 ) ) x ln x .
Then
ln f ( x ) = x ln x [ ln ln ( x 2 ) ln ln ( x 1 ) ] = x ln x [ ln ( 1 + ln ( 1 2 / x ) ln x ) ln ( 1 + ln ( 1 1 / x ) ln x ) ] = x ln x [ ln ( 1 2 x ln x + o ( 1 x ln x ) ) ln ( 1 1 x ln x + o ( 1 x ln x ) ) ] = x ln x [ 2 x ln x + 1 x ln x + o ( 1 x ln x ) ] = 1 + o ( 1 ) .
Therefore lim x + f ( x ) = e 1
Uriah Molina

Uriah Molina

Beginner2022-11-23Added 7 answers

Let L be the desired limit so that
log L = log { lim x ( log ( x 2 ) log ( x 1 ) ) x log x } = lim x log ( log ( x 2 ) log ( x 1 ) ) x log x  (by continuity of log) = lim x x log x log ( log ( x 2 ) log ( x 1 ) ) = lim t 0 + log t t log ( log ( 1 2 t ) log t log ( 1 t ) log t )  (putting  x = 1 / t ) = lim t 0 + log t t log ( 1 log ( 1 t ) log ( 1 2 t ) log ( 1 t ) log t ) = lim t 0 + log t t log ( 1 log ( 1 t ) log ( 1 2 t ) log ( 1 t ) log t ) log ( 1 t ) log ( 1 2 t ) log ( 1 t ) log t ( log ( 1 t ) log ( 1 2 t ) log ( 1 t ) log t ) = lim t 0 + log t t log ( 1 t ) log ( 1 2 t ) log ( 1 t ) log t = lim t 0 + log ( 1 t ) log ( 1 2 t ) t 1 log ( 1 t ) log t 1 = lim t 0 + log ( 1 t ) log ( 1 2 t ) t 1 0 1 = lim t 0 + log ( 1 t 1 2 t ) t = lim t 0 + log ( 1 + t 1 2 t ) t = lim t 0 + log ( 1 + t 1 2 t ) t 1 2 t 1 1 2 t = lim t 0 + 1 1 2 t = 1
and hence L = 1 / e.

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