ANALYSIS: Prove that lim x->c of sqrt{f(x)} = sqrt{L}

[mathmajor2013]

Homework Statement

Suppose that f(x)>=0 in some deleted neighborhood of c, and that lim x->c f(x)=L. Prove that lim x->c sqrt{f(x)}=sqrt{L} under the assumption that L>0.

Homework Equations

When 0<|x-c|<delta, |f(x)-L|<epsilon.

The Attempt at a Solution

When, 0<|x-c|<delta,

|sqrt{f(x)}-sqrt{L}| <= |sqrt{f(x)}-sqrt{L}| |sqrt{f(x)}+sqrt{L}|=|f(x)-L|< epsilon.

I just have a feeling this isn't how I am supposed to do this problem...help would be much appreciated.

 

[SammyS]

Homework Statement

Suppose that f(x)>=0 in some deleted neighborhood of c, and that lim x->c f(x)=L. Prove that lim x->c sqrt{f(x)}=sqrt{L} under the assumption that L>0.

Homework Equations

When 0<|x-c|<delta, |f(x)-L|<epsilon.

The Attempt at a Solution

When, 0<|x-c|<delta,

|sqrt{f(x)}-sqrt{L}| <= |sqrt{f(x)}-sqrt{L}| |sqrt{f(x)}+sqrt{L}|=|f(x)-L|< epsilon.

I just have a feeling this isn't how I am supposed to do this problem...help would be much appreciated.

|sqrt{f(x)}-sqrt{L}| <= |sqrt{f(x)}-sqrt{L}| |sqrt{f(x)}+sqrt{L}| ← is false if |sqrt{f(x)}+sqrt{L}| < 1