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?

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.
