Prove that f(x)=sin(x)/x + cos(sqrt(x)) + 10x isuniformly continuousin her domain.

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- January 3rd 2010, 10:06 AMAlso sprach ZarathustraUniformly continuous #2
Prove that f(x)=sin(x)/x + cos(sqrt(x)) + 10x is

**uniformly continuous**in her domain. - January 3rd 2010, 01:35 PMJose27
It's easy to prove that for differentiable functions it is equivalent:

1) has a bounded derivative

2) is Lipschitz

Since Lipschitz unif. cont. we only have to prove has bounded derivative at infinity which follows almost immediately so is unif. cont. in any interval of the form where , but (I believe, I didn't really check it) that it is not unif. cont. in since the sequence seems (!) to map to an unbounded sequence. - January 3rd 2010, 01:45 PMEnrique2
Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value. - January 3rd 2010, 01:51 PMJose27
- January 3rd 2010, 01:53 PMPlato
- January 3rd 2010, 02:05 PMAlso sprach Zarathustra
And without derivatives?

- January 3rd 2010, 02:13 PMKrizalid
I'm sorry but being analysis questions I think we're allowed to invoke some of stuff from Calculus I, that is what Jose27 just said: bounded derivative implies Lipschitz.

Why making these things harder than they look like. - January 3rd 2010, 02:26 PMAlso sprach Zarathustra
The tools for this questions not include derivatives. Lipschitz yes! Derivatives no!

- January 3rd 2010, 02:32 PMEnrique2
Yes, clearly as Plato and Jose27 have noticed my former argument is not correct, we only have defined the function on positive numbers. The argument works in [0,1] instead of [-1,1] .

The argument that I believe krizalid is suggesting is not so trivial. The bounded derivative argument only works easily outside an interval with a>0. In [0,a] you can use that when approaches 0, but it is not for me easier than using Heine Cantor in a bounded interval and observig that the derivative is bounded outside, that is the correct point to solve a lot of these kind of problems. It was not my idea at all being "too much sophisticated", simply I didn't see the argument in [0,a], that is nice!. - January 3rd 2010, 02:39 PMEnrique2

Sorry! Without the derivatives, proving Lipschitz is harder. Then I believe you will be allowed to use Heine Cantor in [0,a], and making calculations outside, but I don't see at the moment an immediate way for suggesting you how to do these calculations in a simple way, sorry again!