giới hạn HSLG >.< max ngum help

Question

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

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

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

score 5 · Answer 1

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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