极限，趋于正无穷，已知(x^2 1/x-1-αx-β)=1,求α,β？

用洛必达法则求极限 lim [1/(x^2-1) - 1/(x-1)] x→1

先通分,得(xlnx-x+1)/[(x-1)lnx].利用洛必达法则,上下同时求导得lnx/[lnx+(x-1)/x] 再利用洛必达法则,得(1/x)/[(1/x)+(1/x^2)] 于是极限为1/2

limx趋于无穷(1-2/x)^x/2-1的极限?

lim[x→∞] (1+1/x)^(x/2) =lim[x→∞] [(1+1/x)^x]^(1/2) =e^(1/2) =√e 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的＂选为满意回答＂按钮,谢谢.

求极限lim(X趋于正无穷)(1-1/X)^X^1/2

(lnx)^(1/x)=e^[ln((lnx)^(1/x))]=e^[(lnlnx)/x],应用罗必塔法则可知lim(lnlnx)/x=lim(1/(xlnx))=0,因此题目答案为e^0,即1

求极限 lim X无穷大 x^2-1/2x^2-x-1

lim(x→∞) (x^2-1)/ (2x^2-x-1)=lim(x→∞) (x+1)(x-1)/ (2x+1)(x-1)=lim(x→∞) (x+1)/ (2x+1)=lim(x→∞) (1+1/x)/ (2+1/x)=1/2

求x趋向1时[1/(x^2-1)-1/(x-1)]的极限

x趋向1时[1/(x^2-1)-1/(x-1)]的极限=lim(1-(x+1))/(x^2-1)=-limx/(x^2-1)=无穷大

高数极限(x趋于无穷)[x^2/(x^2-1)]^x

[x^2/(x^2-1)]^x=1/[1-1/x^2]^x=1/[1-1/x^2]^[(-x^2)(-1/x)] [1-1/x^2]^(-x^2)趋于e, (-1/x)趋于0 lim(x趋于无穷)[x^2/(x^2-1)]^x=1

(x-1/x^2-x)(1+cosx)求极限,x趋于正无穷,谢谢

先求这部分:(x-1)/(x²-x)=(1-1/x)/(x-1)分子极限为1,分母极限为正无穷,因此极限为0,是无穷小(1+cosx)为有界函数,无穷小与有界函数的乘积为无穷小,因此本题结果是0希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的＂选为满意回答＂按钮,谢谢.

lim趋于无穷 [(x^2-1)/(x^2+1)]^x 求解

解:lim(x->∞){[(x^2-1)/(x^2+1)]^x} =lim(x->∞){[1+(-2)/(x^2+1)]^[((x^2+1)/(-2))*((-2x)/(x^2+1))]} ={lim(x->∞)[(1+(-2)/(x^2+1))^((x^2+1)/(-2))]}^{lim(x->∞)[(-2x)/(x^2+1)]} =e^{lim(x->∞)[(-2x)/(x^2+1)]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e) =e^{lim(x->∞)[(-2/x)/(1+1/x^2)]} (分子分母同除x^2) =e^[0/(1+0)] =1.

求极限:lim(x→1)x/(x^2-1)

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

当x趋近于1时,求(x^2-x+1)/(x-1)^2的极限

解: (x^2-x+1)/(x-1)^2= (x^2-2x+1+x) / (x-1)^2=[(x-1)^2+x] / (x-1)^2=1 + x / (x-1)^2 当x—>1时,(x-1)^2---> 0 由于(x-1)^2≥0恒成立,所以,(x-1)^2是正向趋近于0.