<s>[INST] Evaluating $\lim_{n \to \infty} (1 + 1/n)^{n}$ I was recently thinking about how I could evaluate the famous limit of 'e' as I haven't ever seen a proof. I can't really find anything online so I've tried to evaluate the limit myself. And I was also thinking it would be nonsensical to use L'Hopital's rule, am I right?
So I did the following:
$$\lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = \exp \left(\lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n} \right) \right)$$
$$=\lim_{n \to \infty} \exp \left( n \cdot \left( \frac{1}{n} - \frac{(1/n)^{2}}{2} + \cdots \right) \right)$$
$$= \lim_{n \to \infty} \exp \left( 1 - \frac{(1/n)}{2} + \cdots \right)$$
$$= e$$
I am not sure, is my logic correct or does it create circularity by taking logarithms and assuming $$f(x) = \exp \left( \ln f(x) \right)$$ ?
 [/INST] Here is the "standard" proof:
Let $a_{n}=  \left(1 + \frac{1}{n} \right)^{n}= \frac{(n+1)^n}{n^n}$.
Then
$$\frac{a_{n+1}}{a_n}=\frac{(n+2)^{n+1}}{(n+1)^{n+1}}\frac{n^n}{(n+1)^n}=\frac{(n+2)^{n+1}n^{n+1}}{(n+1)^{2n+2}}\frac{n+1}{n}$$
$$=\left( \frac{n^2+2n}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}=\left( 1-\frac{1}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}$$
Now, by Bernoulli
$$\left( 1-\frac{1}{(n+1)^{2}}\right)^{n+1}\geq 1-\frac{n+1}{(n+1)^{2}}=\frac{n}{n+1}$$
Thus
$$\frac{a_{n+1}}{a_n} \geq \frac{n}{n+1}\frac{n+1}{n}=1 \,.$$
This shows that $a_n$ is increasing.
Now, let 
Let $b_{n}=  \left(1 + \frac{1}{n} \right)^{n+1}= \frac{(n+1)^{n+1}}{n^{n+1}}$.
Then
$$\frac{b_n}{b_{n+1}}=\frac{(n+1)^{n+1}}{n^{n+1}}\frac{(n+1)^{n+2}}{(n+2)^{n+2}}=\left( \frac{(n+1)^2}{n(n+2)} \right)^{n+2}\frac{n}{n+1}$$
$$\geq \left(1+ \frac{n+2}{n(n+2)} \right)\frac{n}{n+1}=1$$
Thus $b_n$ is decreasing.
As $a_n \leq b_n$ it follows that $a_n$ is increasing and bounded from above by $b_1$..
Now, if we denote the limit of this sequence by $e$ (which is the historic definition), it can be proven from 
$$\left(1 + \frac{1}{n} \right)^{n}=\left(1 + \frac{1}{n} \right)^{n+1} =e$$
that $(e^x)'=e^x$.
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