real analysis - Why is $e^{x}$ not uniformly continuous on $mathbb{R}$?

It seems intuitively very clear that $e^{x}$ is not uniformly continuous on $\mathbb{R}$. I'm looking to 'prove' it using $\epsilon$-$\delta$ analysis though. I reason as follows:

Suppose $\epsilon > 0$; in fact, fix it to be $\epsilon=1$.

For contradiction, suppose that $\exists \delta >0$ s.t. $$ (\star) \ |x-y|<\delta \Rightarrow |e^{x}-e^{y}|<\epsilon=1 \text{ for } x,y \in \mathbb{R}.$$
Note that $e^{x+\delta}-e^{x}=e^{x}(e^{\delta}-1)$. So, for $x$ large enough (so that RHS $>1$), the relation $(\star)$ does not hold.

This is our contradiction, and so the exponential function is not uniformly continuous on $\mathbb{R}$.

Is this reasoning correct and sufficient?

Thanks.

Answer

If you are truly looking for a rigorous answer then you need to justify the "So, for $x$ large enough...".

For instance, here is a very rigorous solution along the lines you suggest:

Assume 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 $|e^x-e^y| < 1$. Let $a=\delta/2$. Since $\lim_{x\to\infty }e^x=\infty$ and since $e^a-1>0$ it follows that $\lim_{x\to \infty }e^x(e^a-1)=\infty$. 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.