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Question

$1. \mathop {\lim }\limits_{x \to \pi /3}\frac{tan^{3}x-3tanx}{cos(x+\pi /6)}$

$2.\mathop {\lim }\limits_{x \to \pi /4}(\pi -4x).tan2x$

$3.\mathop {\lim }\limits_{x \to 0}\frac{x^{2}}{\sqrt{1+xsinx}-\sqrt{cosx}}$

Nero · Answer 1 · 4/3/2016 9:55:24 PM

1. Biến đổi tử một chút thì đc $-\frac{sin3x}{cos^3x}$

$L=\frac{-sin3x}{cos^3x.cos(x+\frac{\pi}{6})}$

Đặt

$u=x-\frac{\pi}{3}\Rightarrow u\rightarrow 0$

$L=\frac{sin3u}{cos^3(u+\frac{\pi}{3}).sinu}$ ~

$\frac{3}{cos^3(u+\pi/3)}$ ~

$24$

2. Đặt

$u=x-\frac{\pi}{4}\Rightarrow u\rightarrow 0$

$L=-4u.tan(2u-\pi/2)=-4u.cot2u$ ~

$-2cos2u$ ~

$-2$

3. Liên hợp

$\frac{(\sqrt{1+xsinx}+\sqrt{cosx})x^2}{1+xsinx-cosx}=\frac{\sqrt{1+x^2}+\sqrt{cosx}}{\frac{2sin^2\frac{x}{2}+x^2}{x^2}}$ ~

$4/3$

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