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