I used all the trig and log tricks but still can't compute this limit lim...

Santino Bautista

Santino Bautista

Answered

2022-06-30

I used all the trig and log tricks but still can't compute this limit
lim x 0 ln ( 1 + sin 2 ( 2 x ) ) 1 cos 2 ( x ) .

Answer & Explanation

Bornejecbo

Bornejecbo

Expert

2022-07-01Added 19 answers

Just using the standard limit lim t 0 ln ( 1 + t ) t = 1 you get
ln ( 1 + sin 2 ( 2 x ) ) 1 cos 2 ( x ) = ln ( 1 + sin 2 ( 2 x ) ) sin 2 ( 2 x ) sin 2 ( 2 x ) sin 2 x = 4 cos 2 x ln ( 1 + sin 2 ( 2 x ) ) sin 2 ( 2 x ) x 0 4 1 1 = 4
gnatopoditw

gnatopoditw

Expert

2022-07-02Added 5 answers

We know that (by asymptotic relations):
ln ( 1 + f ( x ) ) f ( x ) x α f ( x ) 0
Also:
sin ( x ) t x 0
And:
cos ( x ) 1 1 2 x 2 x 0
So, your limit is going to be:
lim x 0 ln ( 1 + sin 2 ( 2 x ) ) 1 cos 2 ( x ) lim x 0 sin 2 ( 2 x ) 1 cos 2 ( x ) lim x 0 4 x 2 1 cos 2 ( x ) = lim x 0 4 x 2 ( 1 + cos ( x ) ) ( 1 cos ( x ) ) lim x 0 4 x 2 2 1 2 x 2 = 4

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