Respuesta :

If x → 0, then : ( sin mx ) ~ ( mx ). .............................. (1) __________________________________ 
.. lim (x→0) [ ( 1 - cos 4x ) / ( 1 - cos 6x ) ] 
= lim (x→0) [ ( 2 sin² 2x ) / ( 2 sin² 3x ) ] dx 
= lim (x→0) { [ ( sin 2x )( sin 2x ) ] / [ ( sin 3x )( sin 3x ) ] } 
= lim (x→0) [ (2x) (2x) ] / [ (3x) (3x) ] .............. from (1) 
= lim (x→0) ( 4 / 9 ) 
= 4/9 ........................................... Ans. _____________________________ 
Happy To Help ! _____________________________
apologiabiology
(1-cos(4x))/(1-cos(6x))
l'hopital's rule can be applied because it results in 0/0

take deritiive of top and bottom

(4sin(4x))/(6sin(6x))
if we evaluate
still 0/0
take derititive again

(-16cos(4x))/(-36(cos(6x))
iff we do x=0
-16/-36
16/36
4/9 is the limit

it approaches 4/9

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