Showing a function converges to e

Question

I'm trying to show that the following limit converges to $e$: $$\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = e$$ where $e$ is defined as follows: $$ e = \lim_{n\rightarrow\infty} (1+\frac{1}{n})^n$$ I have the following work done, somewhat showing convergence from the right:

Take $\epsilon > 0$. Since the limit as $n \rightarrow \infty$ exists for $(1+\frac{1}{n})^n$, there exists a positive integer $N \in \mathbb{Z}^+$ such that: $$\left|e - (1+\frac{1}{n})^n\right| < \epsilon $$ for all positive integers $n \geq N$. If we let $\delta = \frac{1}{N}$, then for any real number $x$ satisfying $0 < x< \delta = \frac{1}{N}$, we have $\frac{1}{x} > N$ which implies: $$\left|e - (1 + x)^{\frac{1}{x}}\right| < \epsilon \tag{1}$$ This shows that the right limit of our function tends to $e$ as $x \rightarrow 0$.

I have two question.

1. The same approach doesn't quite work from the left, so I'm stuck on how to approach that part of the question.
2. Since $x$ is not an integer, I'm not sure I can make the final claim, labeled (1). Is there any way to resolve this lapse in rigor?

Edit: I'm not sure I made this clear but the definition of $e$ I use is intended to be sequential. I'm having trouble reconciling the sequential and real definitions of limits.

Answer 1

Hint: Substitute $\dfrac{1}{x}=n$ in your function, and change the limits, and see how after the substitution the function is exactly $e$'s definition.

Answer 2

I'll start from 2 - you can say that for every $x$ there exists $N$ such that $\frac{1}{N}<x$, and so instead of taking $\delta =\frac{1}{N}$, just take $\delta <\frac{1}{N}$.

as for 1, you can use the identity $\lim _{n\rightarrow \infty}(1-\frac{1}{n})^n=\frac{1}{e}$ to solve in a similar way