Find <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-ORD"

shmilybaby4i

shmilybaby4i

Answered question

2022-06-24

Find lim x 0 ( ln ( cos x ) x 1 + x x ) efficiently
Now, it looked to me like a classic L'Hôpital's rule case. Indeed, I used it (twice), but then things became messy and complicated.
Am I missing the point of this exercise? I mean, there must be a "nicer" way. Or should I stick with this road?
EDIT:
Regarding Yiorgos's answer: Why is the following true?
ln ( 1 x 2 2 ) x 2 2

Answer & Explanation

Kaydence Washington

Kaydence Washington

Beginner2022-06-25Added 32 answers

Hints.
I. ln cos x ln ( 1 x 2 2 ) x 2 2
II.
x x + 1 x = x 2 ( x + 1 ) x x 1 + x + x = x ( x + 1 ) 1 1 + x + 1 1 2 .
Devin Anderson

Devin Anderson

Beginner2022-06-26Added 6 answers

By direct computation:
lim x 0 + log ( cos x ) x ( x + 1 1 ) = lim tan x x 2 x + 1 + x + 1 1 = lim sec 2 x x + 1 x / ( 2 x + 1 ) 2 ( x + 1 ) + 1 2 x + 1 = 1 1 / 2 + 1 / 2 = 1
But surely there is a nicer or more elegant method?
I note that only the one-sided limit exists unless of course you are evaluating:
l i m x 0 [ log ( cos x ) x ( x + 1 1 ) ]
However, I digress.

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