Simple absolute value problem with inequalities

[complexhuman]

"Simple" absolute value problem with inequalities

OK...Im totally stuck and could use some help :)
given...for all e>0, d>0...the following holds
|x-a|<d => |f(x) - f(a)| < e
where f(x) = sqrt(x)

how do I find d in terms of e?


Thanks in advance

 

[SGT]

|x - a| = |[f(x) - f(a)][f(x) + f(a)| < e.|f(x) + f(a)| < d
I don't think you can get simpler than that.

 

[EnumaElish]

I am going to assume you meant: for all e > 0 there is some d > 0 such that |$x-a$|< d implies |$\sqrt x - \sqrt a$| < e.

($\sqrt x - \sqrt a$)² < e²

$x + a - 2\sqrt{x a}$ < e²

$x - a + 2(a-\sqrt{x a})$ < e²

$x - a$ < e² - $2(a-\sqrt{x a})$

If $x - a$ > 0 then d = e² - $2(a-\sqrt{x a})$

(So d depends on x, and I guess that's okay.)

I need to think about the case where $x - a$ < 0.

 

[arildno]

It is simplest to note that:
$$|\sqrt{x}-\sqrt{a}|=\frac{|x-a|}{\sqrt{x}+\sqrt{a}}$$
and proceed from there.

 

[complexhuman]

well...I end up with something like $$|x-a|<d => |x-a|=e|\sqrt{x}+\sqrt{a}|$$...And that's where I am stuck on :(

yah...d has to be independent of x...its one of those proving limit typa thing. I am just allowed to assume a = 4 first

 

[HallsofIvy]

If you are attempting to prove that $\sqrt{x}$ is continuous for all positive values of x, then d does not have to be independent of d. That's only true for uniform continuity.

If you have $|x-a| |x-a|=e|\sqrt{x}+\sqrt{a}|$ and x is "sufficiently close to a", say, |x-a|< 1/2, so that a- 1/2< x< a+ 1/2, what can you say about $\sqrt{x}+ \sqrt{a}$?

 

[rsnd]

well...I end up with something like $$|x-a|<d => |x-a|=e|\sqrt{x}+\sqrt{a}|$$...And that's where I am stuck on :(

yah...d has to be independent of x...its one of those proving limit typa thing. I am just allowed to assume a = 4 first

hehe...how did you get $$|x-a|<d => |x-a|=e|\sqrt{x}+\sqrt{a}|$$?
I think it should be $$|x-a|<d => |x-a|<e|\sqrt{x}+\sqrt{a}|$$