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 $$f(x) = \exp \left( \ln f(x) \right)$$ ?
 A: Taking logs is a straightforward approach. I prefer to use a first order expansion based on the fact that $\ln$ is differentiable, rather than a series expansion, but they amount to the same thing here.
If $\phi(x) = \ln (x+1)$, then $\phi(x) = 0 + 1 \cdot x + o(x)$, where $\frac{o(x))}{x} \to 0$, as $x \downarrow 0$. Hence $\lim_{x \downarrow 0} \frac{\phi(x)}{x} = 1$. It follows that $\lim_{n \to \infty} n \ln(1+ \frac{1}{n}) = 1$.
Since $x \mapsto e^x$ is continuous, we have $\lim_{n \to \infty} e^{n \ln(1+ \frac{1}{n})} = \lim_{n \to \infty} (1+\frac{1}{n})^n = e^1 = e$.
A: To avoid circularity, it depends on what you define how, e.g. some authors define the number $e$ by $\lim(1+\frac1n)^n$. Of course, then they need to show that $\exp(1)=e$.
The prettiest (in my opinion) introduction of $e,\exp,\ln$ is by showing that  the space of solutions of the differential equation  $f'(x)=f(x)$ is onedimensional. Then let $\exp$ be the solution with $\exp(0)=1$, immediately obtain $\exp(x+y)=\exp(x)\exp(y)$ (by uniqueness of solution), $\exp(nx)=\exp(x)^n$, $\exp(x)\ne0$, etc. Then we define $e:=\exp(1)$ (because that allows us to use the suggestive notation $e^x$ for $\exp(x)$).
Now in that setup the limit can be obtained without even digging deeper into l'Hospital and $\ln$ (apart from its local existence): Since $\exp'(0)=1\ne0$, the function does have an inverse near $x=0$, hence for $n$ big enough, there is $x_n\approx 0$ such that $\exp(x_n)=1+\frac1n$ (you can even avoid invoking the inverse/implicit function theorem and yous get along with the intermediate value theorem). Since $1=\exp'(0)=\lim_{n\to\infty}\frac{\exp(x_n)-\exp(0)}{x_n}$, we get $nx_n\to 1$ and hence $(1+\frac1n)^n=\exp(nx_n)\to \exp(1)=e$. 
A: 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.
A: 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 $(e^x)'=e^x$.
