lim-x-0-cosx-sinx-

Question Number 1200 by sumitkumar4799@gmail.com last updated on 14/Jul/15

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\mathrm{cos}{x}.\mathrm{sin}{x}\right] \\ $$

Answered by prakash jain last updated on 14/Jul/15

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\right]=\mathrm{0} \\ $$

Answered by Rasheed Soomro last updated on 16/Jul/15

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left[{cos}\:{x}\:.{sin}\:{x}\right]=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({cos}\:{x}\right).\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({sin}\:{x}\right) \\ $$$$\:\:\:\:\:=\mathrm{1}.\mathrm{0}=\mathrm{0} \\ $$$${No}\:{need}\:{to}\:{change}\:{cos}\:{x}.{sin}\:{x}\:{into}\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\:\mathrm{2}{x} \\ $$

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