$f_a(x) = e^{ax}$ is uniformly continuous over $[0, \infty)$? Let $f: \mathbb {R} \rightarrow \mathbb {R}$ defined by $f_a(x) = e^{ax}$.
a) Prove that $f(x) = e^x$ is not uniformly continuous. 
b) Determine for wich values of $a$ the function $f_a(x)$ is uniformly continuous over $[0, \infty)$.
My tried is:
For a) Suppose that $e^x$ is uniformly continuous on $\mathbb {R}$.
Let $\epsilon = 1$. Thus there is $\delta >0$ such that for all $x,y\in \mathbb R$ if $|x-y|<\delta $ then $|e^x-e^y| < 1$, so $a=\delta/2$; since $\lim_{x\to\infty }e^x=\infty$ and since $e^a-1>0$ then $\lim_{x\to \infty }e^x(e^a-1)=\infty$. Consequently, there is some $x\in \mathbb {R}$ such that $e^x(e^a-1)>1$. However, taking $y=x+a$ we have $|x-y|<\delta$ while $|e^x-e^y|=e^x(e^a-1)>1$, a contradiction.
Is that correct?
Can you help me for b) i have problems to determine this values , if $a<1$ i´m not sure if $f_a(x)$ is uniformly continuous, please some help.

If $a \leq 0$, then $f_a(x) = e^{ax}$ is uniformly continuous on $[0,\infty)$. 
Note that
$$|e^{ax}-e^{ay}| = |e^{ax}||e^{a(y-x)}-1| \leq |e^{a(y-x)}-1| < \epsilon,$$
if $|x-y| < \delta$ where $\delta$ does not depend on $x$ or $y$ because $e^{ax}$ is continuous at $x=0$
