Is it this function uniformly continuous?

Question

Is $f(x) = \cos (e^x)$ uniformly continuous on $(0,+\infty)$?

I saw that in $0$ exist and it's limitate the limit of the fuction, so for Haine-Cantor the fuction is uniformly continuous on $[0,M]$. I don't know how to do when $x$ goes to infinite. How can I do?

Thanks.

Answer 1

If $x_{k}=\ln k\pi$ for $k=1,2,\dots$ then $\left|f\left(x_{2k}\right)-f\left(x_{2k+1}\right)\right|=2$ while $\lim_{k\rightarrow\infty}\left|x_{2k}-x_{2k+1}\right|=0$.

This contradicts that $f$ is uniformly continuous on $\left(0,\infty\right)$.

E.g. no $\epsilon>0$ can be found with $|x-y|<\epsilon\Rightarrow |f(x)-f(y)|<1$.

Answer 2

Hint: Prove that, despite $f(x)$ being a twice differentiable and bounded function, $f'(x)$ is unbounded. Using Lagrange's theorem, prove that the uniform continuity of $f(x)$ on $\mathbb{R}^+$ would lead to a contradiction.

Answer 3

Hint: If $f$ is continuous on $(a,b)$ and ‎$‎‎\lim_{x\to a ‎^{+}‎} ‎f(x)‎$‎ and $‎‎‎‎‎\lim_{x\to b ‎^{-}‎} ‎f(x)‎$ exist, ‎then ‎‎$‎f(x)‎$ ‎is ‎uniformly ‎continuous‎

Answer 4

There is a very useful theorem for these type of questions which says that : If the limit of a function exists at +/- infinity and the function is continuous , then the function is uniformly continuous.