lim(cosx)^1 cot^2 x x趋向于0
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原式=lim(x->0)e^[cot²xln(cosx)]=e^[lim(x->0)ln(cosx)/tan²x]=e^[lim(x->0)ln(cosx)/x²]=e^
LIM(趋近与0)(cot^2x-1/x^2)=lim(x^2*cos^2x-sin^2x)/(x^2*sin^2x)=lim[(x^2+1)cos^2x-1]/x^4=lim[(1/2)*(x^2+
原式=lim(x→0){[1+(cosx-1)]^[(1/(cosx-1))(-1)]}=1/lim(x→0){[1+(cosx-1)]^(1/(cosx-1))}=1/lim(t→0)[(1+t)^
x→0时,运用等价无穷小,即1-cosx~x^2/2(1-cosx等价于x^2/2,在乘除中可以直接替换)sinx~x(同理,在乘除中可以直接替换)于是原式=(x^2/2)/(x*x)=1/2
sin2x/[(sinx+cosx-1)(sinx+1-cosx)]=sin2x/[(sinx-(1-cosx)(sinx+1-cosx)]=sin2x/[(sin^2x-(1-cosx)^2]=si
1+cosx=1+2(cosx/2)^2-1=2(cosx/2)^21+cosx-------=2(cosx/2)^2/2sin(x/2)*cos(x/2)=cot(x/2)sinx
一下都省略极限过程x→0设A=lim(cosx+sinx)^1/x,则lnA=limln(cosx+sinx)/x=lim[ln(cosx+sinx)]'/x'【L'Hospital法则】=lim(c
1-pi*pi(x^2-1)/cosx在点x=pi是连续的,所以代入x=pi就是所求的极限值.
lim(1-cosx)x趋向0=1-cos0°=1-1=0
(1+cosx)/(1-cosx)+(1-cosx)/(1+cosx)通分=((1+cosx)^2+(1-cosx)^2)/1-cos^2(x)=2*(1+cos^2(x))/sin^2(x)因为1=
y=(sinx/x)^(cosx/1-cosx)lny=(cosx(lnsinx-lnx)/(1-cosx)limlny=lim(cosx(lnsinx-lnx)/(1-cosx)=lim(lnsin
X趋向0lim(xsinx)/(1-cosx)=X趋向0lim(xsinx)(1+cosx)/(1-cos^2x)=X趋向0limx(1+cosx)/sinx)=X趋向0lim(1+cosx)[x/s
答:lim(x→0)(1-cosx)/x²=lim(x→0)2sin²(x/2)/[4*(x/2)²]=lim(t→0)(1/2)(sint/t)²=1/2
lim(1+x²)^cot²x=lim(1+x²)^(1/x²)(x²cot²x)=lime^(x²/tan²x)=e
原式=lim(x->0){[1+(cosx-1)]^[(1/(cosx-1))(-1)]}=1/lim(x->0){[1+(cosx-1)]^(1/(cosx-1))}=1/lim(t->0)[(1+
1/cosx在x=0处连续,直接代值即可lim(x→0)(1/cosx)=1/cos0=1
x趋近于0时,tanx→x,cotx→1/x,(1+x)^(1/x)→e原式=lim(1+x)^(1-2/x)=lim(1+x)/(1+x)^(2/x)=1/e²