Which is bigger x <mrow class="MJX-TeXAtom-ORD"> l o g (

Brenden Tran

Brenden Tran

Answered question

2022-06-24

Which is bigger x l o g ( x ) or ( l o g ( x ) ) x
I'm trying to find out which is bigger x l o g ( x ) or ( l o g ( x ) ) x as x
I tried to take the log of both but I didnt't reach any where.

Answer & Explanation

Jayce Bates

Jayce Bates

Beginner2022-06-25Added 18 answers

You can try to take lim x + x log ( x ) / log ( x ) x . To do that, see that:
lim x + x log ( x ) log ( x ) x = lim x + exp ( log ( x ) log ( x ) ) exp ( x log ( log ( x ) ) ) = lim x + exp ( log ( x ) log ( x ) x log ( log ( x ) ) ) = exp ( lim x + ( log ( x ) ) 2 x log ( log ( x ) ) )
and so you have to compare ( log ( x ) ) 2 and x log ( log ( x ) ). So take
lim x + ( log ( x ) ) 2 x log ( log ( x ) )
This equals
lim x + x log ( log ( x ) ) ( ( log ( x ) ) 2 x log ( log ( x ) ) 1 )
which is a much easier limit to analyse. Applying l'Hôpital's rule to the fraction inside the parentheses we have
lim x + 2 log ( x ) / x log ( log ( x ) ) + 1 / log ( x )
As x approaches infinity, the numerator approaches 0 and the denominator approaches infinity, so the whole limit approaches 0. Therefore our original limit can be found because
lim x + ( log ( x ) ) 2 x log ( log ( x ) ) = lim x + x log ( log ( x ) ) ( ( log ( x ) ) 2 x log ( log ( x ) ) 1 ) = + ( 1 ) =
Now, this was just the exponent of e in the definition of our original limit. Thus:
lim x + x log ( x ) log ( x ) x = exp ( lim x + ( log ( x ) ) 2 x log ( log ( x ) ) ) = exp ( ) = 0
So we reach the conclusion that, as x + grows faster than x log ( x )
Fletcher Hays

Fletcher Hays

Beginner2022-06-26Added 6 answers

Hint: ln y y is decreasing for y > e

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