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?

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. 
