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Question

1. $\mathop {\lim }\limits_{x \to 0}\frac{1-cos4x}{cos5x-cos3x}$

2. $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1+sinx}-\sqrt{1-sinx}}{x}$

3. $\mathop {\lim }\limits_{x \to 0}\frac{cosx.sinx-tanx}{x^2.sinx}$

Nero · Answer 1 · 4/2/2014 11:50:32 AM

3. $\mathop {\lim }\limits_{x \to 0}\frac{cosxsinx-tanx}{x^2.sinx}$

= $\mathop {\lim }\limits_{x \to 0}\frac{sinx(cos^2x-1)}{x^2.sinx.cosx}$

=$\mathop {\lim }\limits_{x \to 0}\frac{-x^2.(\frac{sinx}{x})^2}{x^2.cosx}$

=$\mathop {\lim }\limits_{x \to 0}\frac{-1}{cosx}=-1$

Sửa 02-04-14 11:50 AM

Nero
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Trả lời 02-04-14 11:48 AM

Nero
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cam on ban – tozakendo 02-04-14 02:15 PM

Nero · Answer 2 · 4/2/2014 11:42:58 AM

2. $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1+sinx}-\sqrt{1-sinx}}{x}$

= $\mathop {\lim }\limits_{x \to 0}\frac{2sinx}{x(\sqrt{1+sinx}+\sqrt{1-sinx})}$

= $\mathop {\lim }\limits_{x \to 0}\frac{2}{\sqrt{1+sinx}+\sqrt{1-sinx}}=1$

Trả lời 02-04-14 11:42 AM

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Nero · Answer 3 · 4/2/2014 11:38:43 AM

1. $\mathop {\lim }\limits_{x \to 0}\frac{1-cos4x}{cos5x-cos3x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{2sin^22x}{-2sin4x.sinx}$

=$\mathop {\lim }\limits_{x \to 0}\frac{4x^2.(\frac{sin2x}{2x})^2}{-4x^2.\frac{sin4x}{4x}.\frac{sinx}{x}}$

=$ -1$

Trả lời 02-04-14 11:38 AM

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