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Question

a)

$\mathop {\lim }\limits_{x \to 0}\frac{1-\cos 6x}{x^{2}}$

b)

$\mathop {\lim }\limits_{x \to 0}\frac{1-\cos 3x}{1-\cos 5x}$

c)

$\mathop {\lim }\limits_{x \to \frac{\Pi }{2}}\frac{\cos x}{x-\frac{\Pi }{2}}$

d)

$\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{2x+1}-\sqrt[3]{x^{2}+1}}{\sin x}$

Nero · Answer 1 · 4/4/2014 8:57:18 PM

Bình chọn tăng 3 Bình chọn giảm

c.

$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{cosx}{x-\frac{\pi}{2}}$

=

$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{sin(\frac{\pi}{2}-x)}{-(\frac{\pi}{2}-x)}=-1$

Trả lời 04-04-14 08:57 PM

Nero
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299K 319K

Nero · Answer 2 · 4/4/2014 9:14:18 PM

Bình chọn tăng 3 Bình chọn giảm

b.

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

=

$\mathop {\lim }\limits_{x \to 0}\frac{2sin^2\frac{3x}{2}}{2sin^2\frac{5x}{2}}$

=

$\mathop {\lim }\limits_{x \to 0}\frac{(\frac{3x}{2})^2.(\frac{sin\frac{3x}{2}}{\frac{3x}{2}})^2}{(\frac{5x}{2})^2.(\frac{sin\frac{5x}{2}}{\frac{5x}{2}})^2}=\frac{9}{25}$

lịch sử

Sửa 04-04-14 09:14 PM

Nero
96 1 2

299K 319K

Trả lời 04-04-14 08:55 PM

Nero
96 1 2

299K 319K

xem câu d đúng chưa ông anh – Nero 04-04-14 09:24 PM

cai nay sai dap so sửa đi – Dép Lê Con Nhà Quê 04-04-14 09:08 PM

Nero · Answer 3 · 4/4/2014 8:51:11 PM

Bình chọn tăng 3 Bình chọn giảm

a.

$\mathop {\lim }\limits_{x \to 0}\frac{1-cos6x}{x^2}$

=

$\mathop {\lim }\limits_{x \to 0}\frac{2sin^23x}{x^2}$

=

$\mathop {\lim }\limits_{x \to 0}\frac{2.(3x)^2.(\frac{sin3x}{3x})^2}{x^2}=18$

Trả lời 04-04-14 08:51 PM

Nero
96 1 2

299K 319K

Nero · Answer 4 · 4/4/2014 9:24:17 PM

d.

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

=

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

=

$\mathop {\lim }\limits_{x \to 0}\frac{2x}{sinx(\sqrt{2x+1}+1)}-\mathop {\lim }\limits_{x \to 0}\frac{x.x}{sinx[(\sqrt[3]{x^2+1})^2+\sqrt[3]{x^2+1}+1)]}$

=

$\mathop {\lim }\limits_{x \to 0}\frac{2}{\sqrt{2x+1}+1}-\mathop {\lim }\limits_{x \to 0}\frac{x}{(\sqrt[3]{x^2+1})^2+\sqrt[2]{x^2+1}+1}$

=

$1$