analysis, uniform continuouty Have this in one of my finals practice questions:
Given $f(x) = e^{\sin x}$, prove $f(x)$ is uniformly continuous on $\mathbb R$.
The direction I'm thinking of is choosing $x_1 = 2\pi k + \delta$, and $x_2 = 2\pi k + \frac{\delta}2$.
so I need to prove that for all $\varepsilon > 0$, there exists $\delta > 0$ such that $|x_1-x_2|< \delta$ gives $|f(x_1)-f(x_2)|< \varepsilon$.
So, my choice of $x_1$ and $x_2$ gives $|x_1-x_2| = \delta/2 < \delta$,
and $|f(x_1)-f(x_2)| = |e^{\sin(2\pi k + \delta)} - e^{\sin(2\pi k+ \delta/2)}|$ 
This stage is a little dodgy for me.. any help on what to do next would be great.
 A: Since $|e^{\sin(x)}|\leq e$, we have
$$|e^{\sin(x)}-e^{\sin(y)}|= |e^{\sin(x)}||e^{[\sin(y)-\sin(x)]}-1|\leq e|e^{[\sin(y)-\sin(x)]}-1|.$$
Since $e^x$ is continuous at $x=0$, for any $\epsilon >0$ there exists $\delta(\epsilon) >0 $ independent of $x$, such that if $|x|<\delta(\epsilon)$ then $|e^x-1|< \epsilon.$
Consequently, if
$$|\sin(y)-\sin(x)|<\delta_0=\delta(\epsilon/e)$$
then 
$$|e^{[\sin(y)-\sin(x)]}-1|< \epsilon/e.$$
But
$$|\sin(y)-\sin(x)|=2\left|\sin\frac{(y-x)}{2}\right|\left|\cos\frac{(y+x)}{2}\right|\leq2\left|\sin\frac{(y-x)}{2}\right|\leq|y-x|.$$
Therefore, if $|y-x| < \delta_0$ then $|e^{\sin(x)}-e^{\sin(y)}|<\epsilon$ for all $x,y \in \mathbb{R}$.  Since $\delta_0$ does not depend on $x$ and $y$, the function is uniformly continuous. 
A: One approach would be to say that, because the function is periodic with period $2 \pi$, a modulus of continuity at $x \in [0,2\pi]$ (that is, the $\delta=\delta(x,\varepsilon)$ so that $|x-y| < \delta \Rightarrow |f(x)-f(y)|<\varepsilon$) also works at $x+2k\pi$ for any $k \in \mathbb{Z}$. Then you could use the usual result about uniform continuity on compact sets.
A: Note that
$$
|F'(x)| = |e^{\sin(x)}\cos(x)| \leq e
$$
So try using the Mean-Value Theorem.
