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Question

a) $\mathop {\lim }\limits_{x \to 0}\frac{x}{\sin x}$

b) $\mathop {\lim }\limits_{x \to 0}(\frac{1}{\sin x}-\frac{1}{\sin 3x})\frac{1}{x}$

c) $\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{\cos x}}{\tan ^{2}x}$

Nero · Answer 1 · 4/5/2014 12:19:32 PM

c. $\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}$

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

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

=$\mathop {\lim }\limits_{x \to 0}(\frac{x}{sinx})^2.\frac{(\frac{1}{4}).(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2cos^2x}{1+\sqrt{cosx}}=\frac{1}{8}$

Sửa 05-04-14 12:19 PM

Nero
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Trả lời 04-04-14 08:33 PM

Nero
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Nero · Answer 2 · 4/4/2014 8:30:24 PM

a. Ta có công thức : $\mathop {\lim }\limits_{x \to 0}\frac{sinnx}{nx}=1$ hoặc $\mathop {\lim }\limits_{x \to 0}\frac{nx}{sinnx}=1$

Nên $\mathop {\lim }\limits_{x \to 0}\frac{x}{sinx}=1$

Trả lời 04-04-14 08:30 PM

Nero
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Nero · Answer 3 · 4/4/2014 9:03:41 PM

b. $\mathop {\lim }\limits_{x \to 0}\frac{1}{x}(\frac{sin3x-sinx}{sinx.sin3x})$

=$\mathop {\lim }\limits_{x \to 0}\frac{2sinx.cos2x}{x.sinx.sin3x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{2cos2x}{x.sin3x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{2.3x.cos2x}{x.sin3x}=\mathop {\lim }\limits_{x \to 0}\frac{2cos2x}{x}=0$

Trả lời 04-04-14 09:03 PM

Nero
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299K 309K

câu này giải sai rồi .. nếu làm vậy thế 0 vào sẽ ra vô định – kutequata_vu 17-04-14 12:43 AM

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