Prove that f(x)=sin(x)/x + cos(sqrt(x)) + 10x is uniformly continuous in her domain.
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.
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.
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!.
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!