Question Number 1200 by sumitkumar4799@gmail.com last updated on 14/Jul/15
![lim_(x→0) [cosx.sinx]](Q1200.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\mathrm{cos}{x}.\mathrm{sin}{x}\right] \\ $$
Answered by prakash jain last updated on 14/Jul/15
![lim_(x→0) [(1/2)sin 2x]=0](Q1203.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\right]=\mathrm{0} \\ $$
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](Q1225.png)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left[{cos}\:{x}\:.{sin}\:{x}\right]=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({cos}\:{x}\right).\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({sin}\:{x}\right) \\ $$$$\:\:\:\:\:=\mathrm{1}.\mathrm{0}=\mathrm{0} \\ $$$${No}\:{need}\:{to}\:{change}\:{cos}\:{x}.{sin}\:{x}\:{into}\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\:\mathrm{2}{x} \\ $$