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1.lim

1.lim x→1 [(1/(1-x)-2/(1-x^2)] 解:原式=lim x→1 [(1+x)/(1-x^2)-2/(1-x^2)] =lim x→1[(x-1)/(1-x^2)] = lim x→1[-1/(1+x)] =-1/2 2.lim n→无限 n-[(n^2-2n+7)/n+1] 解:原式=lim n→无限 (n^2+n/n+1)-[(n^2-2n+7)/n+1] =lim n→无限(3n-7)/(n+1...

极限是不存在的。分子里有因式x-1,那极限一定是存在的。在看这道题,因为分母趋近于0,分子趋近于一个常数,当然整体趋近于无穷。但在分子里没有因式x-1时,也不能说极限不存在,如lim(x→1)sin(x-1)/(x-1)=1,lim(x→1)tan(x-1)/(x-1)=1.lim(x→1)...

lim [x/(1+x)]²ˣ x→∞ =lim {[1+ 1/(-x-1)]⁻ˣ⁻¹}⁻²·[1+ 1/(-x-1)]⁻² x→∞ =e⁻²·(1+0)⁻² =e⁻²·1 =e⁻²

用平方差公式化简: (有问题欢迎追问@_@)

lim[x-->1][1/(1-x)-3/(1-x^3)] =lim[x-->1][(1+x+x^2-3)/(1-x^3)] =lim[x-->1][(x^2+x-2)/(1-x^3)] =lim[x-->1][(x-1)(x+1)+(x-1)]/[(1-x)(1+x+x^2)] =-lim[x-->1][(x+2)/(1+x+x^2)] =-1

这个利用了重要极限,lim x趋于无穷 (1+1/x)的x次方=e,大括号里面只是配成中重要极限,外面的-1是因为n比-n

先讲求lim(x→∞)(1+1/X)^f(x)转化为“e^limf(x)”的方法 lim(x→∞)(1+1/x)^x=e总是存在的, 如果极限lim(x→∞)f(x)/x=b也存在, 则lim(x→∞)(1+1/x)^f(x) = lim(x→∞)[(1+1/x)^x]^[f(x)/x] =lim(x→∞)[(1+1/x)^x]^[lim(x→∞)f(x)/x)] =e^[lim(x→∞)f(x)/x)] ...

没有说明n趋向0或n趋向∞,那两种情况都写吧: lim[n→0] (1-1/n)^(n²) =e^lim[n→0] n²ln(1-1/n),令n=1/y,∴n²=1/y² =e^lim[y→∞] ln(1-y)/y²,用洛必达法则 =e^lim[y→∞] [-1/(1-y)]/(2y) =e^(-1/2)lim[y→∞] 1/[y(1-y)],分母...

令x=-1/t lim(1-x)^1/x x->0 = lim1/ [(1+1/t)^t] t->∞ =1/e

是x趋于1吗 令x=1+1/a 则a趋于无穷 1/(1-x)=-a 所以原式=lim(1+1/a)^(-a) =1/lim(1+1/a)^a =1/e

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