Question: I'm curious what is the solution of this. Is this just same as ordinary natural log?

$\lim \limits\_{n\to∞ }(1+\frac{1}{n} + \frac{1}{n^2})^n =e?$

A few people says the $\frac{1}{n^2}$ just goes to $'0'$, so it's same as $'e'$.
But why? Why $\frac{1}{n}$ remains meaningful, while $\frac{1}{n^2}$ goes to zero??

I'm asking this because I'm stuck while deriving a formula.
I post the part of deriving as a picture. Look. From 1st, they go to 3rd line, by using 2nd. And I assume this suppose $\lim \limits\_{n\to∞ }(1+\frac{1}{n} + \frac{1}{n^2})^n $ is just same as 'e'.

Thank you genius![enter image description here](https://i.stack.imgur.com/o06D4.png)

Answer: 
Remember that $(1+\epsilon)^n\simeq 1+\epsilon n$ for $\epsilon\to 0$. If $\epsilon\sim O(1/n)$ then $\epsilon n \sim O(1)$ so such a term is relevant. If $\epsilon \sim O(1/n^2)$ then $\epsilon n\sim O(1/n)\to 0$.

Hence if you have $\left(1 + \frac{1}{n}+\frac{1}{n^2}\right)$ you have something in the form $(1+\epsilon\_1 +\epsilon\_2)^n$ with $\epsilon\_1\sim O(1/n)$ and $\epsilon\_2 \sim O(1/n^2)$. So you'll have $1+\epsilon\_1 n + \epsilon\_2 n$. The term in $\epsilon\_1$ is of order $1$ (actually it's $e-1$) and the other is of order $1/n$ and becomes negligible.

To see things more rigorously you can take the log...
$$\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n = e^{n\log(1+1/n+1/n^2)}=e^{(1+O(1/n)) +(1/n+O(1/n^2))}=e^{1+O(1/n)}\to e$$
$$\begin{align}
\lim\_{n\to \infty} \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n &=e^{\lim\_{n\to\infty} n\left(\frac{1}{n}+\frac{1}{n^2}\right)}\\
&=e^{\lim\_{n\to\infty} \left(1+\frac{1}{n}\right)} \\
&=e
\end{align}$$