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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...

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

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...

因为无法上传图解,请参看本人中心的专门解答:

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

无意中看到这道题.大家也有几种解法了: [(x+1)/(x+2)]-[1/(x+7)]=[(x+2)/(x+3)]-[1/(x+6)] 为简化起见,特设:(x+1)=a,(x+2)=b,(x+3)=c,(x+6)=d,(x+7)=e 于是公式成为:a/b-1/e=b/c-1/d 先移项:a/b-b/c=1/e-1/d 等式两边同乘以:bcde,也就是同乘(x+2...

lim(1+1/x+1)^x =lim(1+1/x+1)^[(x+1)-1] =[lim(1+1/x+1)^(x+1)]^(-1) =e^-1 =1/e

函数应该这样吧:1/(x+1)-3/(x^3+1) lim [1/(x+1)-3/(x^3+1)] = lim [ (x^3-3x-2)/(x+1)(x^3+1)] =lim ( 3x^2-3)/[(x^3+1)+(x+1)3x^2](罗比达) =lim [( 3x^2-3)/(4x^3+3x^2+1)](罗比达) =lim[6x/(12x^2+6x)] =-6/6=-1

lim[n→+∞][1/n²+1/(n + 1)² + 1/(n + 2)² + ... + 1/(n + n)²] = lim[n→+∞]{1/[n(1+0/n)]²+1/[n(1 + 1/n)]² + 1/[n(1 + 2/n)]² + ... + 1/[n(1 + n/n)]²} = lim[n→+∞] (1/n²)[1+1/(1 + 1/n)²...

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