TITLE: Evaluating $\lim_{n \to \infty} (1 + 1/n)^{n}$
QUESTION [5 upvotes]: 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 $$f(x) = \exp \left( \ln f(x) \right)$$ ?

REPLY [2 votes]: There are several ways to approach this question.

In this answer, it is shown that
$$
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^\infty\frac{x^n}{n!}
$$
which gives the usual value for $e^x$, in particular, $e=e^1$.

Note that
$$
a_n=\left(1+\frac1n\right)^n\lt\left(1+\frac1n\right)^{n+1}=b_n
$$
In this answer, it is shown that $a_n$ is an increasing sequence and $b_n$ is a decreasing sequence. Thus, $a_n$ is an increasing sequence bounded above by $a_n\lt b_n\lt b_1=4$ and $b_n$ is a decreasing sequence bounded below by $2=a_1\lt a_n\lt b_n$. Thus, both sequences converge.
Since
$$
\lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}1+\frac1n=1
$$
the limits of both sequences are the same. Call that limit $e$, then we have
$$
2=a_1\lt\lim_{n\to\infty}a_n=e=\lim_{n\to\infty}b_n\lt b_1=4
$$
By computing $a_n$ and $b_n$ for large $n$, we can bound the value of $e$ as closely as we want.