Question

解决当x→2时，y=x²→4， 问δ等于多少，使当／x-2／＜δ时，／y-4／＜0.001

解决当x→2时，y=x²→4， 问δ等于多少，使当／x-2／＜δ时，／y-4／＜0.001？？

Answer 1

y=x^2

/y-4/<0.001

/x^2-4/<0.001

-0.001<x^2-4<0.001

4-0.001<x^2<4+0.001

3.999<x^2<4.001

1.9997<x<2.0002or-2.0002<x<-1.9997

-0.0003<x-2<0.0002or       -4.0002<  x-2<-3.9997

(-0.0003,0.0002)or(-4.0002,-3.9997)

a>/x-2/,a>/x-2/max,

y=/x-2/,

令t=x-2

y=/t/

t:(-0.0003,0.0002)or(-4.0002,-3.9997)

=(-0.0003,0)u[0,0.0002)u(-4.0002,-3.9997)

1. 当t(-0.0003,0)u(-4.0002,-3.9997)

   在R上是减函数，tmin=-4.0002,ymax=-(-4.0002)=4.0002

2. 时，t<0,y=-t,

t=-4.0002取不到，所以ymax=4.0002取不到,在t(-0.0003,0)上的值域为（0,0.0003），在(-4.0002,-3.9997)上的值域为(3.9997,4.0002)

y属于（0,0.0003）u[0,0.0002)u(3.9997,4.0002)=[0,0.0003)U(3.9997,4.0002)

y<4.0002

tmax=0,ymin=-0=0,t=0取不到，所以ymin=0取不到，y>ymin=0,y>0

2.t属于[0,0.0002),y=/t/=t

t=0.0002,ymax=0.0002,t=0.0002取不到，则ymax=0.0002取不到，y<0.0002

t=0,ymin=0,

y；[0,0.0003)U(3.9997,4.0002)

a>y恒成立，a>=4.0002

a的范围是[4.0002,+无穷）