[Toán 11] Topic giới hạn

[vivietnam]

C1:
[TEX]\lim_{x \to 0}\frac{(1+2mx+m^2x^2)^n-(1+2nx+n^2x^2)^m}{x^2[(1+mx)^n+(1+nx)^m]}=\lim_{x \to 0}\frac{1+2mnx+m^2nx^2-1-2mnx-n^2mx^2}{MT}=\lim_{x \to 0}\frac{m^2n-n^2m}{(1+mx)^n+(1+nx)^m}=\frac{m^2n-n^2m}{2}[/TEX]
C2:
lên đại học sẽ biết

 

[vivietnam]

c,[TEX]\lim_{\frac{1}{x} \to 0}(sin(ln(x+1))-sin(lnx))[/TEX]

còn bài này
nhìn vậy thui chứ dễ lắm
mọi người làm thử đi

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

tiếp tục,thử mấy bài này coi,khá hay

c,[TEX]\lim_{\frac{1}{x} \to 0}(sin(ln(x+1))-sin(lnx))[/TEX]

do

mặt khác

là hàm tuần hoàn nên không tồn tại giới hạn

 

[bigbang195]

b,[TEX]\lim_{\frac{1}{x} \to 0}(\sqrt[3]{x^3+x^2-1}-x)[/TEX]

 

[duynhan1]

Xét tử :

[TEX](1+mx)^n=C_n^k (mx)^k \ \ (k= 0 \rightarrow n) [/TEX]

[TEX](1+nx)^m= C_m^q (nx)^q \ \ (q= 0 \rightarrow m)[/TEX]

[TEX]\Rightarrow (1+mx)^n - (1+nx)^m = (C_n^0 - C_m^0) + (C_n^1.mx-C_m^1.nx) + (C_n^2 (mx)^2 -C_m^2 (nx)^2)+ x^3(.............) =(C_n^2 (mx)^2 -C_m^2 (nx)^2)+ x^3(.............) [/TEX]

[TEX]\Rightarrow \lim_{x\to\0}\frac{(1+mx)^n-(1+nx)^m}{x^2}= C_n^2.m^2 - C_m^2.n^2= \frac{n(n-1)m^2 - m(m-1).n^2}{2} = \frac{mn(n-m)}{2}[/TEX]

Sao nó lại ngược dấu so với anh vivietnam :-s

 

[duynhan1]

[TEX]\huge \lim_{x \to \frac{\pi}{3}} \frac{sin x - \sqrt{3}cosx}{sin3x} =\lim_{x \to \frac{\pi}{3}} \frac{\frac12 sin (x -\frac{\pi}{3})}{-sin3(x-\frac{\pi}{3})} =\lim_{x \to \frac{\pi}{3}} \frac{-1}{2.(3 - 4 sin^2(x -\frac{\pi}{3}))} = - \frac16 [/TEX]

[TEX]\huge \lim_{x%20\to%20\pi}%20\frac{\cos%20x+1}{\sin%203x} = \lim_{x%20\to%20\pi}%20\frac{sin^2x}{(1-cosx) \sin%203x} =\lim_{x%20\to%20\pi} \frac{sinx}{(1-cosx)(3-4sin^2x)} = 0 [/TEX]

Tương tự :

[TEX]\huge \lim_{x%20\to%20\frac{\pi}{2}}\bigg(\frac{1}{cosx}-\tan%20x\bigg)= \lim_{x%20\to%20\frac{\pi}{2}}\bigg(\frac{cos^2x}{cos x(1+ sinx)}\bigg) = 0[/TEX]

 

[duynhan1]

[TEX]\huge \lim_{x%20\to%200}\frac{\sqrt{3}-\sqrt{2+\cos%202x}}{1-\cos%20x} [/TEX]

[TEX]\huge = \lim_{x%20\to%200} \frac{(1+cosx)( 1- cos 2x)}{sin^2x(\sqrt{3}+\sqrt{2+\cos%202x})} = \lim_{x%20\to%200} \frac{2(1+cosx)}{(\sqrt{3}+\sqrt{2+\cos%202x})} = 2(\sqrt{3}-\sqrt{2}) [/TEX]

[TEX]\huge \lim_{x%20\to%200}\frac{1-\cos%202x\cos%203x\cos%204x}{x^2} = \lim_{x%20\to%200}\frac{2-cos^23x - cos^2x(4cos^2x - 3 )}{2x^2} =\lim_{x%20\to%200}\frac{1+sin^2 3x -(1-sin^2x)(1-4sin^2x)}{2x^2}\\ =\lim_{x%20\to%200}\frac{1+sin^2 3x - 4sin^4x + 5sin^2x}{2x^2} = =\lim_{x%20\to%200}(\frac{sin 3x}{3x})^2 . \frac{9}{2}- \frac{sin^2x}{x^2} . 2sin^2x + \frac52(\frac{sin^2x}{x^2})= \frac92+\frac52=7 [/TEX]

:-s

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[bigbang195]

 

[ngomaithuy93]

[TEX] \lim_{x\to\pi} \frac{sin(\pi cosx)}{tanx+cosx}=0[/TEX]