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.
 A: Your proof looks good to me. There are only a few technicalities:


*

*You should consider an arbitrary $\delta>0$ because the contrapositive of
$$\forall \epsilon>0,\ \exists\delta>0,\ \forall|x-y|<\delta,\ |f(x)-f(y)|<\epsilon$$
is
$$\exists \epsilon>0,\ \forall\delta>0,\ \exists|x-y|<\delta,\ |f(x)-f(y)|\ge\epsilon.$$

*You should consider something like $e^{x+\delta/2}-e^x$ rather than $e^{x+\delta}-e^x$ because $|(x+\delta)-x|$ is not smaller than $\delta$.

*As Clayton pointed out in the comment section, if you claim that $(\star)$ does not hold when $x$ is large enough, you'd better explicitly write down an instance of $x$.

A: 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. 
A: Another way to prove that $e^{x}$ is not uniformly continuous on $\mathbb{R}$ using an $\varepsilon$-$\delta$ argument is to consider a fairly rigid consequence of this class of functions by means of the following result:
If $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous then there exists constants $A, B >0$ such that $|f(x)|< A|x| +B$ $ \forall x \in \mathbb{R}$
Proof: Take $\varepsilon =1$ in the definition of uniform continuity so that there exists a $\delta >0$ such that if $x, y \in \mathbb{R}$ satisfy $|x-y|< \delta $, then $|f(x)-f(y)|< 1.$ In particular, setting $y=0$ and using the triangle inequality, we get that $|x|< \delta \Rightarrow |f(x)|< |f(0)|+1 < |f(0)|+1 + |x| = A_{1}|x| + B_{1}$, where $A_{1}=1$ and $B_{1}= |f(0)|+1.$
Now suppose $|x| \geq \delta$ and WLOG let $x >0$ (the case of $x<0$ uses a similar argument.) Set $n = \lfloor \frac{2x}{\delta}\rfloor$ so that $0 \leq x - \frac{n\delta}{2} \leq \frac{\delta}{2}.$ From $0,$ we then approach $x$ in $n$ steps of $\frac{\delta}{2}$ and use uniform continuity on the intervals $(\frac{i\delta}{2}, \frac{(i+1)\delta}{2})$as follows:
$\begin{align}|f(x)-f(0)|&\leq \displaystyle \sum_{i=0}^{n-1}\left|f\left(\frac{i\delta}{2}\right)- f\left(\frac{(i+1)\delta}{2}\right)\right|+ \left|f\left(\frac{n \delta}{2}\right)- f(x)\right| \\&\leq n+1\\&= \frac{2x}{\delta} +1\end{align}$
$\Rightarrow |f(x)|< |f(0)|+1+\frac{2x}{\delta}$
Repeating this argument for the $x\leq -\delta$ case and setting $A= \max\{A_{1}, \frac{\delta}{2}\}$ and $B = B_{1}$, the proof is complete.
Using this result, we can immediately see that $f(x)= e^x$ is not uniformly continuous or else we would have $e^x < A|x| +B$, which is an obvious contradiction for large $x.$ This argument also shows that non-linear polynomials and other fast-growing functions like $\tan(x)$ are not uniformly continuous on $\mathbb{R}.$ 
A: On the contrary, let us assume that $e^{x}$ is uniformly continuous on $\mathbb{R}$. So by
def, given $\epsilon > 0$, $\exists \delta > 0$, such that $|e^{x}-e^{y}| < \epsilon$ 
whenever $|x-y| < \delta$. 
Now let $x=n+\frac{\delta}{2}$ and $y=n$, clearly x and y are close enough so that,
$|e^{x}-e^{y}| < \epsilon$
by mean value theorem we got,
$e^{\zeta}.|x-y|=|e^{x}-e^{y}| < \epsilon$,  where $y < \zeta < x$
as $e^{x}$ is increasing function we must have $e^{n} = e^{y} < e^{\zeta}$, hence
$e^{n} . \frac{\delta}{2} < e^{\zeta}|x-y| =|e^{x}-e^{y}| < \epsilon$
$e^{n}<\frac{2.\epsilon}{\delta}$, which is a contradiction as $e^{n}$ tends to infinity as n tend.
