- #1
fyziky
- 4
- 0
Hello all,
My question is as follows:
f:[1,[tex]\infty[/tex]) is defined by f(x)=[tex]\sqrt{x}[/tex]+2x (1[tex]\leq[/tex]x<[tex]\infty[/tex]) Given [tex]\epsilon[/tex]>0 find [tex]\delta[/tex]>0 such that if |x-y|<[tex]\delta[/tex] then |f(x)-f(y)|<[tex]\epsilon[/tex]
It seems I am being asked to show continuity, and not uniform continuity, so my approach is this, but I am not sure it works:
|f(x)-f(y)|=[tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex] + 2(x-y) after minor manipulation
since [tex]\sqrt{x}[/tex] function is uniformly continuous it is safe to set [tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex]<[tex]\delta[/tex]/2 and defining |x-y|<[tex]\delta[/tex]/4 we get
[tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex] + 2(x-y) [tex]\leq[/tex] [tex]\delta[/tex]/2 + 2[tex]\delta[/tex]/4=[tex]\delta[/tex]
since [tex]\delta[/tex] can equal [tex]\epsilon[/tex] we just need make |x-y|<[tex]\epsilon[/tex]/4
any help would be appreciated, I am not sure i have the right idea here. thank you.
My question is as follows:
f:[1,[tex]\infty[/tex]) is defined by f(x)=[tex]\sqrt{x}[/tex]+2x (1[tex]\leq[/tex]x<[tex]\infty[/tex]) Given [tex]\epsilon[/tex]>0 find [tex]\delta[/tex]>0 such that if |x-y|<[tex]\delta[/tex] then |f(x)-f(y)|<[tex]\epsilon[/tex]
It seems I am being asked to show continuity, and not uniform continuity, so my approach is this, but I am not sure it works:
|f(x)-f(y)|=[tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex] + 2(x-y) after minor manipulation
since [tex]\sqrt{x}[/tex] function is uniformly continuous it is safe to set [tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex]<[tex]\delta[/tex]/2 and defining |x-y|<[tex]\delta[/tex]/4 we get
[tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex] + 2(x-y) [tex]\leq[/tex] [tex]\delta[/tex]/2 + 2[tex]\delta[/tex]/4=[tex]\delta[/tex]
since [tex]\delta[/tex] can equal [tex]\epsilon[/tex] we just need make |x-y|<[tex]\epsilon[/tex]/4
any help would be appreciated, I am not sure i have the right idea here. thank you.