已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/12 23:00:55
已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
lim(n^2+1/n+1+an-b)
=lim(n^2+1/n+1+(an-b)(n+1)/n+1)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
又 lim(n^2+1/n+1+an-b)=1,可知:
[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)中[(1+a)n^2的系数1+a=0,即a=-1
故:lim(n^2+1/n+1+an-b)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
=lim[(a-b)n+(1-b)]/ (n+1)=1
则有:a-b=1,即b=a-1=-2
故:a=-1,b=-2
=lim(n^2+1/n+1+(an-b)(n+1)/n+1)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
又 lim(n^2+1/n+1+an-b)=1,可知:
[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)中[(1+a)n^2的系数1+a=0,即a=-1
故:lim(n^2+1/n+1+an-b)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
=lim[(a-b)n+(1-b)]/ (n+1)=1
则有:a-b=1,即b=a-1=-2
故:a=-1,b=-2
已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
已知:lim (n→∞) [(n^2+n)/(n+1)-an-b]=1 ,求a,b的值
已知lim[(an^2+5n-2)/(3n+1)-n]=b,求a,b的值
已知lim(n2+1/n+1-an-b)=1求a,b 对了 那是n的平方 n趋近正无穷
lim[(n^2+1)/(n+1)-an-b]=o求a和b的值 lim[1/(a-1)^n]=0 求a的范围
看清题,已知lim((n平方+1/n+1)-an-b)=1,求实数a,b的值.
已知lim((an2+5n-2)/(3n+1) -n)=b 求a b的值
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
已知极限等式lim(n→∞)[(n平方+1/n+1)-an-b]=1,则ab的值为?
已知lim[(3n^2+cn+1)/(an^2+bn)-4n]=5,求常数a、b、c的值
lim(n2+2n+2)/(n+1)-an)=b,求a,b
lim (n→∞) (n^2/(an+b)-n^3/(2n^2-1))=1/4 求a,b