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

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- Jan 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. - Jan 3rd 2010, 01:35 PMJose27
It's easy to prove that for differentiable functions it is equivalent:

1)$\displaystyle f$ has a bounded derivative

2)$\displaystyle f$ is Lipschitz

Since Lipschitz $\displaystyle \Rightarrow$ unif. cont. we only have to prove $\displaystyle f$ has bounded derivative at infinity which follows almost immediately so $\displaystyle f$ is unif. cont. in any interval of the form $\displaystyle [a, \infty )$ where $\displaystyle a>0$, but (I believe, I didn't really check it) that it is not unif. cont. in $\displaystyle (0, \infty )$ since the sequence $\displaystyle 1/n$ seems (!) to map to an unbounded sequence. - Jan 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. - Jan 3rd 2010, 01:51 PMJose27
- Jan 3rd 2010, 01:53 PMPlato
- Jan 3rd 2010, 02:05 PMAlso sprach Zarathustra
And without derivatives?

- Jan 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. - Jan 3rd 2010, 02:26 PMAlso sprach Zarathustra
The tools for this questions not include derivatives. Lipschitz yes! Derivatives no!

- Jan 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 $\displaystyle [0,a]$ with a>0. In [0,a] you can use that $\displaystyle sin(\sqrt{x})=O( \sqrt{x})$ when $\displaystyle x$ 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!. - Jan 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!