Prove an inequality using only a limit I have to show that if $x\le y,$ $(y-x)e^x\le e^y-e^x\le(y-x)e^y$, then use this to show that $\lim_{n\to\infty}e^{-n}=0$.  All I am given is that $\lim_{n\to\infty}(1+x/n)^n=e^x$
First I showed that if $x\le y$, then $e^x\le e^y$ simply by plugging into the given limit. I am not sure if this even helps.
 A: I'm going to assume that all you know about the function I'll relabel 
$$e(x)=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$
is that the limit exists for all $x$.  In that case, here's a start on what you want to show:  If $x\le y$, then, as soon as $n\gt |y|$ (which we need so that $0\lt1+{x\over n}\lt1+{y\over n}$ even when $x$ and $y$ are negative), we have
$$\begin{align}
\left(1+{y\over n}\right)^n&-\left(1+{x\over n}\right)^n\\
&=\left[\left(1+{y\over n}\right)-\left(1+{x\over n}\right)\right]\left[\left(1+{y\over n}\right)^{n-1}+\left(1+{y\over n}\right)^{n-2}\left(1+{x\over n}\right)+\cdots+\left(1+{x\over n}\right)^{n-1} \right]\\
&={y-x\over n}\left[\left(1+{y\over n}\right)^{n-1}+\left(1+{y\over n}\right)^{n-2}\left(1+{x\over n}\right)+\cdots+\left(1+{x\over n}\right)^{n-1} \right]\\
&\le{y-x\over n}\left[\left(1+{y\over n}\right)^{n-1}+\left(1+{y\over n}\right)^{n-2}\left(1+{y\over n}\right)+\cdots+\left(1+{y\over n}\right)^{n-1} \right]\\
&=(y-x)\left(1+{y\over n}\right)^{n-1}\\
&=(y-x)\left(1+{y\over n}\right)^n\left(1+{y\over n}\right)^{-1}
\end{align}$$
At this point taking limits as $n\to\infty$ gives
$$e(y)-e(x)\le(y-x)e(y)$$
The other inequality, $(y-x)e(x)\le e(y)-e(x)$, can presumably be handled in similar fashion.  I'm a bit less sure how to approach the corollary $\lim_{n\to\infty}e^{-n}=0$.
A: Hint:$e^y - e^x = e^{x}\left(e^{y-x}-1\right)$, you need to prove: $\forall n \geq 1, \left(1+\frac{y-x}{n}\right)^n - 1 \geq y-x$, and this is evidently true by Bernoulli's inequality. The other direction is similarly done. To get the limit $0$, put $y = n, x = 0$ into the left inequality: $n=(n-0)e^{0} \leq e^n - e^{0} = e^n - 1\to e^n \geq n+1\to 0<e^{-n} \leq \dfrac{1}{n+1}\to \displaystyle \lim_{n\to \infty} e^{-n} = 0$ by Squeeze's theorem.
To show that: $x \leq y \to e^x \leq e^y$. Let $t = y-x \geq 0$, and observe that:
$\left(1+\dfrac{t}{n}\right) \geq 1, \forall t \geq 0 \to \left(1+\dfrac{t}{n}\right)^n \geq 1 \to \displaystyle \lim_{n\to \infty} \left(1+\frac{t}{n}\right)^n \geq 1 \to e^{t} \geq 1 \to e^{y-x} \geq 1 \to e^y \geq e^x$
A: Once you showed that $f(x)=e^x$ is increasing, you can use the mean value theorem (because $f$ is differentiable over $\mathbb{R}$): for $x\leq y$, $e^y-e^x=e^z(y-x)$, for some $z\in [x,y]$. Using the fact that $f$ is increasing you easily get
$$
(y-x)e^x\leq (y-x)e^z= e^y-e^x \leq (y-x) e^y.
$$
