Limit of a sequence that looks like $e$ How does one go about calculating the limit of the sequence
$$a_n = \left(1 + \dfrac{1}{n^2} \right)^n.$$ I understand the cases of 
$
b_n = \left(\alpha + \dfrac{\beta}{n}\right)^n,
$
in terms of $e$, but I am missing the "trick" with regards to $a_n$.
 A: Since $\left(1+\dfrac{1}{n^{2}}\right)^{n^{2}}\rightarrow e$, then $n^{2}\log\left(1+\dfrac{1}{n^{2}}\right)\rightarrow\log e=1$. Now
\begin{align*}
n\log\left(1+\dfrac{1}{n^{2}}\right)=\dfrac{1}{n}\cdot n^{2}\log\left(1+\dfrac{1}{n^{2}}\right)\rightarrow 0\cdot 1=0,
\end{align*}
so
\begin{align*}
\left(1+\dfrac{1}{n^{2}}\right)^{n}\rightarrow e^{0}=1.
\end{align*}
A: Note that
$$1\le\left(1 + \dfrac{1}{n^2} \right)^n=\left[\left(1 + \dfrac{1}{n^2} \right)^{n^2}\right]^\frac1n\le e^\frac1n$$
thus for squeeze theorem
$$\left(1 + \dfrac{1}{n^2} \right)^n\to 1$$
A: Hint:
$$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}=e$$
A: As long as you know the binomial theorem, you don't need to know anything about $e$:
$$\begin{align}
1\le\left(1+{1\over n^2}\right)^n
&=1+{n\over n^2}+{{n\choose2}\over n^4}+\cdots+{{n\choose n}\over n^{2n}}\\
&\le1+{1\over n}+{n^2\over n^4}+\cdots+{n^n\over n^{2n}}\\
&\le1+{1\over n}+{1\over n^2}+\cdots+{1\over n^n}+\cdots\\
&={1\over1-{1\over n}}\to1
\end{align}$$
A: 
I thought it might be of interest to present a way forward that relies only on Bernoulli's Theorem and the squeeze theorem.  To that end we proceed.


This is as simple as $1$, $2$, and $3$.
First, we note that 
$$\left(1+\frac{1}{n^2}\right)\le \frac1{\left(1-\frac1{n^2}\right)}\tag 1$$

Second, applying Bernoulli's inequality ($(1+x)^n\ge 1+x$ for $x>-1$) to both sides of $(1)$ reveals
$$1+\frac1n\le \left(1+\frac{1}{n^2}\right)^n\le \frac{1}{1-\frac1n}\tag2$$

Third, and finally, applying the squeeze theorem to $(1)$ yields the coveted limit
$$\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^n=1$$

And we are done!
A: You have obtains $$ \lim_{m\rightarrow \infty}\left(1+\frac{1}{m}\right)^m =e.$$
In your case, you have that $m=n^2$, then
$$ \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2}\right)^n=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^{n^2\cdot\frac{1}{n} }=\left[ \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^{n^2 } \right]^{\frac{1}{n}}=e^{\lim_{n\rightarrow \infty}\frac{1}{n}}=e^0=1.$$
