Is $f(x)= \cos(e^x)$ uniformly continuous? As in the topic, my quest is to check (and prove) whether the given function $$f(x)= \cos(e^x)$$is uniformly continuous  on $\left\{\begin{matrix}x \in (-\infty;0]
\\ 
x \in [0; +\infty)
\end{matrix}\right.$ . 
My problem is that I have absolutely no idea how to do it. Any hints will be appreciated and do not feel offended if I ask you a question which you consider stupid, but such are the beginnings. Thank you in advance.

There is uniform continuity for $x \in (-\infty,0]$, because the derivative $f'(x)=-e^x \sin(e^x)$ is bounded there ($|f'(x)| \le 1$ there):
Let $\varepsilon>0$ be given, and suppose $|x_1-x_2|<\delta=\varepsilon$. Apply Lagrange's mean value theorem: $|f(x_1)-f(x_2)|=|f'(\xi)| |x_1-x_2| \le 1 \cdot |x_1-x_2|<\varepsilon$. ($\xi$ is some number between $x_1$ and $x_2$).
There is no uniform continuity on $[0,+\infty)$ though. To see this look at the pairs of points $x_n=\ln (2\pi n),y_n=\ln((2n+1) \pi)$, they get closer than any $\delta$, yet the distance between their f-value is always 2.
