Proving $e^x \sin x$ is not uniformly continuous on $[0,\infty)$ I have to prove that $e^x \sin x$ is not uniformly continuous on $[0,\infty)$ using the mean value theorem. I proved that $\sin x + \cos x \geq 1$ for all $x\in [2 \pi k, 2\pi k + \pi/2], k\in\mathbb{N}$, and I see that $f'(x)=e^x (\sin x + \cos x)\geq e^x$, but I just can't think of way proving this with MVT. Any suggestions?

If $x,y \in [2 \pi k,2 \pi k+\pi/2]$, then the mean value theorem and the inequalities you've written imply that $|f(x)-f(y)| \geq e^{2 \pi k}|x-y|$. So if you pick $\varepsilon=1$ and let $\delta>0$ be arbitrary, then you can choose $k$ large enough that $e^{2 \pi k} \delta > 1$. This line of thinking should give you your result.
