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Question-132007




Question Number 132007 by abdullahquwatan last updated on 10/Feb/21
Answered by bramlexs22 last updated on 10/Feb/21
 lim_(x→0)  ((cos 2x+sin 3x(cos x−cos 2x)−cos x)/(sin^2 x+cos 2x−1))  = lim_(x→0)  (((cos 2x−cos x)(1−sin 3x))/(cos 2x−cos^2 x))  = lim_(x→0)  (((2cos^2 x−cos x−1))/(cos^2 x−1 )) ×lim_(x→0)  (1−sin 3x)  = 1 × lim_(x→0)  (((2cos x+1)(cos x−1))/((cos x+1)(cos x−1)))  = (3/2)
limx0cos2x+sin3x(cosxcos2x)cosxsin2x+cos2x1=limx0(cos2xcosx)(1sin3x)cos2xcos2x=limx0(2cos2xcosx1)cos2x1×limx0(1sin3x)=1×limx0(2cosx+1)(cosx1)(cosx+1)(cosx1)=32
Commented by abdullahquwatan last updated on 10/Feb/21
thanks
thanks

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