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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 im being asked to show continuity, and not uniform continuity, so my approach is this, but im 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, im not sure i have the right idea here. thank you.

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# Homework Help: Epsilon, delta convergence

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