I wanted to prove <munder> <mo form="prefix">lim <mrow class="MJX-TeXAtom-ORD"> x

sedeln5w 2022-06-11 Answered
I wanted to prove lim x ln x x = 0 by the squeeze theorem.
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Answers (2)

hofyonlines5
Answered 2022-06-12 Author has 12 answers
ln x = 2 ln x 1 2 2 x (using the inequality y ln y for y > 0 that you obtained.)
So it follows that: for all x > 0
| ln x x | 2 x x
It follows by squeeze theorem that lim x | ln x x | = 0. Now note that | ln x x | ln x x | ln x x | and hence the result follows by squeeze theorem.
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Hailie Blevins
Answered 2022-06-13 Author has 8 answers
Extending the approximation
e x 1 + x
into
e x 1 + x + x 2 2 > ( x + 1 ) 2 2
and
e x 1 + x + x 2 2 + x 3 6 > ( x + 1 ) 3 6
allows
e x > ( x + 1 ) n n !
for all n Z
So,
x n ln ( 1 + x ) ln n ! n ln x ln n !
So
x n + ln n ! n ln x
1 n + ln n ! n x ln x x
As we can make both terms on the LHS arbitrarily small, we are done.
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