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lim-x-0-cosx-sinx-




Question Number 1200 by sumitkumar4799@gmail.com last updated on 14/Jul/15
lim_(x→0)   [cosx.sinx]
limx0[cosx.sinx]
Answered by prakash jain last updated on 14/Jul/15
lim_(x→0)  [(1/2)sin 2x]=0
limx0[12sin2x]=0
Answered by Rasheed Soomro last updated on 16/Jul/15
lim_(x→0)  [cos x .sin x]=lim_(x→0) (cos x).lim_(x→0) (sin x)       =1.0=0  No need to change cos x.sin x into (1/2)sin 2x
limx0[cosx.sinx]=limx0(cosx).limx0(sinx)=1.0=0Noneedtochangecosx.sinxinto12sin2x

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