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[sandy_26]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!

ĐĂNG BÀI NGAY để cùng trao đổi với các thành viên siêu nhiệt tình & dễ thương trên diễn đàn.

1, [TEX]\lim_{x\to pi/4}[/TEX] ( sin x - cosx)/ cos2x
2, [TEX]\lim_{x\to pi}[/TEX]([TEX]\sqrt{1- tanx}[/TEX] - [TEX]\sqrt{1+tan x}[/TEX] )/ sin 2x
3, [tex] \lim_{x\to 0}[/tex](cos (a+x) -cos(a-x)) / x
4,[tex] \lim_{x\to 0}[/tex] ([TEX]\sqrt{1+sin x}[/TEX] - [TEX]\sqrt{1- sin x}[/TEX]) / tan x
5, [tex] \lim_{x\to 0}[/tex] ([TEX]\sqrt{1 +x. sinx}[/TEX] - [TEX]\sqrt{cos 2x}[/TEX]) / [TEX]x^2[/TEX]
6, [tex] \lim_{x\to 0}[/tex] (sinx. sin2x.....sinnx) / [TEX]x^n[/TEX]
7, [tex] \lim_{x\to +\infty[/tex] ( cosx/2.cosx/4.cosx/8.....cosx/[TEX]2^n[/TEX])

>->->->-

 

[tuyn]

1, [TEX]\lim_{x\to pi/4}[/TEX] ( sin x - cosx)/ cos2x

[TEX] \lim_{x\to \frac{\pi}{4}} \frac{sinx-cosx}{cos^2x-sin^2x}[/TEX]
[TEX]= \lim_{x\to \frac{\pi}{4}} \frac{-1}{sinx+cosx}=-\frac{1}{\sqrt{2}}[/TEX]

2, [TEX]\lim_{x\to pi}[/TEX]([TEX]\sqrt{1- tanx}[/TEX] - [TEX]\sqrt{1+tan x}[/TEX] )/ sin 2x

[TEX] \lim_{x\to \pi} \frac{-2tanx}{sin2x[\sqrt{1-tanx}+\sqrt{1+tanx}]}[/TEX]
[TEX]\lim_{x \to \pi} \frac{1}{cos^2x(\sqrt{1+tanx}+\sqrt{1-tanx})}=\frac{-1}{2}[/TEX]

3, [tex] \lim_{x\to 0}[/tex](cos (a+x) -cos(a-x)) / x

[TEX]cos(a+x)-cos(a-x)=-2sinasinx[/TEX]
[TEX]\Rightarrow \lim_{x \to 0} \frac{-2sinasinx}{x}=-2sina[/TEX]

4,[tex] \lim_{x\to 0}[/tex] ([TEX]\sqrt{1+sin x}[/TEX] - [TEX]\sqrt{1- sin x}[/TEX]) / tan x

[TEX]=\lim_{x \to 0} \frac{2sinx}{tanx(\sqrt{1-sinx}+\sqrt{1+sinx})}[/TEX]
[TEX]=\lim_{x \to 0}\frac{2cosx}{\sqrt{1+sinx}+\sqrt{1-sinx}}=1[/TEX]

5, [tex] \lim_{x\to 0}[/tex] ([TEX]\sqrt{1 +x. sinx}[/TEX] - [TEX]\sqrt{cos 2x}[/TEX]) / [TEX]x^2[/TEX]

[TEX]=lim_{x \to 0} \frac{1+xsinx-cos2x}{x^2(\sqrt{1+xsinx}+\sqrt{cos2x})}[/TEX]
[TEX]=\lim_{x \to 0} \frac{sin^2x+xsinx}{x^2(\sqrt{1+xsinx}+\sqrt{cos2x})}[/TEX]
[TEX]=\lim_{x \to 0} \frac{sinx}{x}. (1+\frac{sinx}{x}). \frac{1}{\sqrt{1+xsinx}+\sqrt{cos2x}}=1[/TEX]

6, [tex] \lim_{x\to 0}[/tex] (sinx. sin2x.....sinnx) / [TEX]x^n[/TEX]

[TEX]=\lim_{x \to 0}\frac{sinx}{x} .\frac{sin2x}{2x}....\frac{sinnx}{nx}.(1.2.3...n)=n![/TEX]

7, [tex] \lim_{x\to +\infty[/tex] ( cosx/2.cosx/4.cosx/8.....cosx/[TEX]2^n[/TEX])

Ta có:
[TEX]cos(\frac{x}{2})cos(\frac{x}{2^2])...cos(\frac{x}{2^n})=\frac{1}{2^nsin(\frac{x}{2^n})}2^ncos(\frac{x}{2})cos(\frac{x}{2^2})...sin( \frac{x}{2^n})cos(\frac{x}{2^n})[/TEX]
[TEX]=\frac{1}{2^nsin(\frac{x}{2})}2^{n-1}cos(\frac{x}{2})cos(\frac{x}{2^2})...cos(\frac{x}{2^{n-1}})sin(\frac{x}{2^{n-1}})[/TEX]
[TEX]=\frac{1}{2^nsin(\frac{x}{2})}2^{n-2}cos(\frac{x}{2})cos(\frac{x}{2^2})...cos(\frac{x}{2^{n-2}})sin(\frac{x}{2^{n-2}})=...=\frac{sinx}{sin(\frac{x}{2^n})}[/TEX]
[TEX]\Rightarrow lim=\infty[/TEX]

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