giới hạn logarit

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nguyenbahiep1

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lim(x->0) lncosx/lncos2x

[laTEX]\lim_{ u \to 0}\frac{ln(1+u)}{u} = 1 ,\lim_{ u \to 0}\frac{sinu}{u} = 1 \\ \\ \lim_{ x \to 0}\frac{ln(1-2sin^2(\frac{x}{2}))}{ln(1-2sin^2x} \\ \\ \lim_{ x \to 0}\frac{ln(1-2sin^2(\frac{x}{2}))}{-2sin^2(\frac{x}{2})}.\frac{-2sin^2x}{ln(1-2sin^2x)}.\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}.\frac{x^2}{sin^2x}.\frac{1}{4} = \frac{1}{4}[/laTEX]
 
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