Prove that $x\mapsto e^x$ is continuous at $x_0 = 1$ ($\delta-\varepsilon$ proof) Prove that the function $$f(x)=e^x:=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ is continuous at $x_0=1$ using the delta epsilon definition of continuous, which is:
$$\forall \varepsilon >0 \exists \delta>0 (\forall x\in D:|x-x_0|<\delta) |f(x)-f(x_0)|<\varepsilon$$
Since, in this particular context, the limit inside $e^x$ was proved to be convergent for all $x$ using the Bernoulli inequality and monotone convergence theorem, I'm struggling to see how I can apply a delta epsilon proof here.  In my limited experience, I've applied it to nothing more than simple algebraic functions, but this is pretty significantly different.  How do I start?
 A: You proved previously that the limit is convergent. This means
$$\forall\epsilon>0\;\exists N\in\mathbb{N}\;{s.t.}\;n\geq N\implies\left|e^x-\left(1-\frac{x}{n}\right)^n\right|<\epsilon$$
To prove continuity of the defined function, we want to show
$$\left|f(x)-f(x_0)\right|=\left|\lim_{n\to\infty}\left(1-\frac{x}{n}\right)^n - \lim_{n\to\infty}\left(1-\frac{x_0}{n}\right)^n\right|<\epsilon$$
Our approach will be to reduce this to
$$\left|e^x-e^{x_0}\right|<\epsilon$$
which is a well known and easy to prove continuous function. Let's start by choosing $N$ so that the sequence is convergent. Now let's cleverly add zero.
$$\begin{align}\left|\left(1-\frac{x}{n}\right)^n - e^x + e^x - \left(1-\frac{x_0}{n}\right)^n\right|&\leq\left|e^x-\left(1-\frac{x}{n}\right)^n\right|+\left|e^x-\left(1-\frac{x_0}{n}\right)^n\right|\\&<\frac{\epsilon}{3}+\left|e^x-\left(1-\frac{x_0}{n}\right)^n\right|\\
&=\frac{\epsilon}{3}+\left|e^x-e^{x_0}+e^{x_0}-\left(1-\frac{x_0}{n}\right)^n\right|\\
&<\frac{2\epsilon}{3}+\left|e^x-e^{x_0}\right|\end{align}$$
We want to choose a $\delta$ such that
$$\left|e^x-e^{x_0}\right|\leq\frac{\epsilon}{3}$$
Consider the case that $x\geq x_0$, then consider $e^x-e^{x_0}=e^{x_0}\left(e^{x-x_0}-1\right)$. So we can choose a $\delta$ such that
$$e^{x_0}\left(e^\delta-1\right)=\frac{\epsilon}{3}\implies\delta=\ln\left(1+e^{-x_0}\frac{\epsilon}{3}\right)$$
which is what we wanted to show. The same can be shown for $x_0>x$. Note that this $\delta$ is dependent on $x_0$ indicating point-wise continuity.
Edit: Fixed based on Jason DeVito's counterexample.
A: Here is some hints that will help you. It's just write it with more details.
Let $x_0\in\mathbb R$ and $\epsilon>0$ be arbitrary real numbers.
Notice that
$$|e^x - e^{x_0}| = |e^x - (1+x/n)^n + (1+x/n)^n - (1+{x_0}/n)^n + (1+{x_0}/n)^n - e^{x_0}| \leq |e^x-(1+x/n)^n| + |(1+x/n)^n-(1+{x_0}/n)^n| + |(1+{x_0}/n)^n-e^{x_0}|$$
so a $\epsilon/3$-argument would be a nice idea.
Of course $|e^x-(1+x/n)^n|$ and $|(1+{x_0}/n)^n-e^{x_0}|$ get small enough ($<\epsilon/3$) as $n$ goes to $\infty$.
So it's sufficient to show that $|(1+x/n)^n-(1+{x_0}/n)^n|$ get small enough ($<\epsilon/3$) as $x$ goes to $x_0$.
Well, here we can use the fact that for a fixed $n$ the function $z\mapsto(1+z/n)^n$ is continuous at $x$ (it is a polynomial).
Hope it helps.
A: Since you already covered the hard part of showing that the limit of sequence $(1+(x/n))^{n}$ exists for all real $x$, the continuity of the function defined via this limit will not seem that difficult to you.
Establish that $e^{x+y} =e^xe^y$ and then note that $$|e^x-e|=e|e^{x-1}-1|$$ and then use the inequality (you need to prove this too) $$1+x\leq e^x\leq \frac{1}{1-x}$$ for $0<x<1$. This should help you to prove that $e^x$ is continuous at $1$ by finding an appropriate $\delta$.
Let me know if you have difficulty proceeding in this manner and then I will provide more details. As a bonus that inequality mentioned above is fundamental and powerful and can be used to prove that $e^x$ is differentiable everywhere with the derivative equal to itself. 
A: I am going to initially write $\exp(x)$ instead of $e^x$ to lesssen the tempatation to use exponential rules, until we show $\exp(x)$ really is an exponential function.  (I am assuming you have already proved some exponential laws in your class).
Assuming $x > 0$, Set $m = n/x$, so that $m\rightarrow \infty$ as $n\rightarrow \infty$.  Substituting this in gives  $\exp(x):=\lim_{n\rightarrow \infty} (1 + x/n)^n = \lim_{m\rightarrow \infty} (1 + 1/m)^{mx} = [\lim_{m\rightarrow \infty} (1+1/m)^m]^x = \exp(1)^x$.  Defining $e:=\exp(1)$, this proves $\exp(x) = e^x$, so it really is an exponential.
We next claim that if $e^x$ is continuous at $x_0 = 0$, then $e^x$ is continuous everywhere.  To see this, choose $x_0\in\mathbb{R}$ suppose $\epsilon>0$ is given.  Then $|e^x - e^{x_0}| = |e^{x_0}(e^{x-x_0} - 1)| = e^{x_0}|e^{x-x_0} - 1|$.  By hypothesis, there is a $\delta$ with the property that if $|y|<\delta$, then $|e^y - 1| < \frac{\epsilon}{e^{x_0}}$.  Then using this $\delta$, if $y=|x-x_0| < \delta$, then $|e^x - e^{x_0}| = e^{x_0}|e^{x-x_0} - 1|   < e^{x_0}\frac{\epsilon}{e^{x_0}} = \epsilon$, as desired.
We next claim that if $e^x$ is continuous from the right at $x_0$, then it is automatically continuous from the left at $x_0$.  The point is that $e^{x_0}\neq 0$, so $1/e^x $ is continuous from the right at $x_0$.  But $1/e^x = e^{-x}$, so this is the same as being continuous from the left.
Thus, we need only show that $e^x$ is continuous from the right when $x_0=0$.  So, let $\epsilon > 0$ be given.  Now, pick $\delta < \min\{1, \frac{\epsilon}{\epsilon+1}\}$.
Assume $0<x<\delta$.  Then using the binomial theorem, \begin{align*} |e^x - 1| &= |\lim_{n\rightarrow\infty} (1+x/n)^n -1| \\ &= \left|\lim_{n\rightarrow\infty} \sum_{k=0}^n {n\choose k}\left( \frac{x}{n} \right)^k -1\right|\\ &= \left| \lim_{n\rightarrow\infty}\sum_{k=1}^n {n\choose k}\frac{1}{n^k} x^k\right| \end{align*}.
We need to estimate ${n\choose k} \frac{1}{n^k}$.  We get \begin{align*} {n\choose k}\frac{1}{n^k} = \frac{\overbrace{n(n-1)...(n-(k+1))}^{k\ terms}}{k!} \frac{1}{n^k} \\ &\leq \frac{n(n)...(n)}{k!} \frac{1}{n^k} \\ &= \frac{1}{k!}\\ &\leq 1\end{align*}
In particular, since $x> 0$, $$\left| \lim_{n\rightarrow\infty}\sum_{k=1}^n {n\choose k}\frac{1}{n^k} x^k\right|\leq \left|\sum_{k=1}^\infty x^k\right| = \left|\frac{x}{1-x}\right|,$$ where we use the geometric series formula (which is valid since $|x|<\delta < 1$.)
Now, one easily verifies that $\frac{x}{1-x} < \epsilon$ iff $x< \frac{\epsilon}{1+\epsilon}$, and since $|x|<\delta < \frac{\epsilon}{1+\epsilon}$.  This completes the proof that $f(x) = e^x$ is continuous.
