Evaluate the limit of ln(cos2x)/ln(cos3x) as x->0

Gardiolo0j

Gardiolo0j

Answered question

2022-09-30

Evaluate the limit of ln ( cos 2 x ) / ln ( cos 3 x ) as x 0
Evaluate Limits
lim x 0 ln ( cos ( 2 x ) ) ln ( cos ( 3 x ) )
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by = L H R . LHR stands for L'Hôpital Rule)
lim x 0 ( ln ( cos ( 2 x ) ) ln ( cos ( 3 x ) ) ) = L H R = lim x 0 ( ln ( cos ( 2 x ) ) ln ( cos ( 3 x ) ) ) = lim x 0 ( ln ( cos ( 2 x ) ) ) ( ln ( cos ( 3 x ) ) ) = lim x 0 ( 2 sin ( 2 x ) cos ( 2 x ) 3 sin ( 3 x ) cos ( 3 x ) ) = lim x 0 ( 2 sin ( 2 x ) cos ( 3 x ) 3 cos ( 2 x ) sin ( 3 x ) ) = L H R lim x 0 ( 2 sin ( 2 x ) cos ( 3 x ) ) ( 3 cos ( 2 x ) sin ( 3 x ) ) = lim x 0 ( 4 cos ( 2 x ) cos ( 3 x ) 6 sin ( 3 x ) sin ( 2 x ) 9 cos ( 3 x ) cos ( 2 x ) 6 sin ( 2 x ) sin ( 3 x ) ) = lim x 0 ( 2 ( 2 cos ( 2 x ) cos ( 3 x ) 3 sin ( 3 x ) sin ( 2 x ) ) 3 ( 3 cos ( 2 x ) cos ( 3 x ) 2 sin ( 3 x ) sin ( 2 x ) ) ) = lim x 0 ( 2 ( 2 cos ( 2 x ) cos ( 3 x ) 3 sin ( 3 x ) sin ( 2 x ) ) ) lim x 0 ( 3 ( 3 cos ( 2 x ) cos ( 3 x ) 2 sin ( 3 x ) sin ( 2 x ) ) ) = 4 9
Could we do it in others ways?

Answer & Explanation

Rachael Singh

Rachael Singh

Beginner2022-10-01Added 4 answers

If you know lim x 0 ln ( 1 + x ) x = 1, then
lim x 0 ln ( cos 2 x ) ln ( cos 3 x ) = lim x 0 ln ( 1 + cos 2 x 1 ) cos 2 x 1 cos 3 x 1 ln ( 1 + cos 3 x 1 ) cos 2 x 1 cos 3 x 1
We have lim x 0 ln ( 1 + cos 2 x 1 ) cos 2 x 1 = 1, lim x 0 ln ( 1 + cos 3 x 1 ) cos 3 x 1 = 1
and lim x 0 cos 2 x 1 cos 3 x 1 = 4 9 , since cos x = 1 x 2 2 + o ( x 3 ) when x 0. We can also use L'Hopital's rule here
overrated3245w

overrated3245w

Beginner2022-10-02Added 3 answers

Once you apply L'Hopital's Rule once, and simplify you get lim x 0 2 sin ( 2 x ) cos ( 2 x ) 3 sin ( 3 x ) cos ( 3 x ) = lim x 0 2 tan 2 x 3 tan 3 x
Now, applying L'Hopital's Rule one more time gives you lim x 0 4 sec 2 2 x 9 sec 2 3 x = 4 9
Alternatively, lim x 0 2 sin ( 2 x ) cos ( 2 x ) 3 sin ( 3 x ) cos ( 3 x ) = lim x 0 2 3 cos 3 x cos 2 x sin 2 x x sin 3 x x = lim x 0 4 9 cos 3 x cos 2 x sin 2 x 2 x sin 3 x 3 x
which is easily broken up into well known limits.
Note: More often than not, it is better to simplify a limit before applying L'Hopital's Rule.

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