Question

当x趋向于正无穷时，y=(x^2-1)/(x^2+3)趋向于1，问X等于多少，则当绝对值x大于X时，绝对值y-1小于0.01?

Answer 1

Y=(X^2-1)/(X^2+3)
Y=(X^2+3)-4/(X^2+3)=(X^2+3)/(X^2+3)+(-4)/(X^2+3)=1-4/(X^2+3)
Y-1=-4/(X^2+3)
绝对值Y-1=4/(X^2+3)<0.01
因为X^2+3>=3，所以4/(X^2+3)<0.01
4<0.01(X^2+3)
X^2>397
X>根号下397或X<-根号下397
因为绝对值X大于X，而当X>根号下397时，X等于它本身。所以得X<- 根号下397。

Answer 2

Y=(X^2-1)/(X^2+3)
Y=(X^2+3)-4/(X^2+3)=(X^2+3)/(X^2+3)+(-4)/(X^2+3)=1-4/(X^2+3)
Y-1=-4/(X^2+3)
绝对值Y-1=4/(X^2+3)<0.01
祝你学业有成