# Question of proof of analysis

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

I tried to work on lsqrt{f(x)}-sqrt(L)l=lsqrt{f(x)}-sqrt(L)llsqrt{f(x)}+sqrt(L)l/lsqrt{f(x)}+sqrt(L)l
=lsqrt{f(x)}-sqrt(L)l/(sqrt{f(x)}+sqrt(L))

But I don't know how to go from here, I'm not sure if it's a correct start as well.

Any help would be appreciated. Thanks!

No, you calculated wrong. What is

$$\frac{|\sqrt{f(x)}-\sqrt{L}||\sqrt{f(x)}+\sqrt{L}|}{ |\sqrt{f(x)}+\sqrt{L}|}$$

I actually don't know what it is...I just don't know how to proceed. Could you help me how I get to start with this question?

SammyS
Staff Emeritus
Homework Helper
Gold Member
You have a deleted neighborhood of c, so isn't this lim x ➙ c , not lim x ➙ a ?

Yes, you're right.

Since f(x) > 0, R>0, Hence 0 < |√(f(x)) - √(R)| < |√(f(x)) + √(R)|.

for ε^2 given there exists δ>0 such that when |x-c| <δ, |f(x) - R| < ε^2.

then

|√(f(x)) - √(R)|^2 < |√(f(x)) - √(R)||√(f(x)) + √(R)| = |f(x) - R| < ε^2,

taking square roots give the desired result.

Thank you for your help. But, sorry, how can I conclude this? Why can I say that lim x->a sqrt{f(x)}=sqrt{R}?

SammyS
Staff Emeritus
Homework Helper
Gold Member
(a - b)(a + b) = a2 - b2 .

What is $|\sqrt{f(x)}-\sqrt{L}||\sqrt{f(x)}+\sqrt{L}|\,?$

Now can you answer micromass's question?

SammyS

Sorry, which part you're asking? Is it for what Trevor Vadas said?

I'm now just wondering how I can conclude this problem from |√(f(x)) - √(R)|^2 < |√(f(x)) - √(R)||√(f(x)) + √(R)| = |f(x) - R| < ε^2.