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: 
Hint: For any $\epsilon >0$ we have $1+\frac 1 n \leq 1+\frac 1 n+\frac 1 {n^{2}} \leq 1+\frac {1+\epsilon} n$ for $n$ sufficiently large. Apply logarithm, multiply by $n$ and take the limit.
$$\left(1+\frac{1}{n}+\frac{1}{n^2}\right)=\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n^2+n}\right) $$
so if $\left(1+\frac{1}{n}\right)^n\to e$, then $\left(1+\frac{1}{n^2+n}\right)^n\to 1$ and $\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n\to e$.