Finding the Limit of a Sequence (No L'Hopital's Rule) Okay, I feel almost silly for asking this, but I've been on it for a good hour and a half. I need to find:
$$\lim_{n \to\infty}\left(1-\frac{1}{n^{2}}\right)^{n}$$
But I just can't seem to figure it out. I know its pretty easy using L'Hopital's rule, and I can "see", that it's going to be $1$, but apparently it's possible to compute using only the results I've proved so far. That is, the sandwich theorem, the addition/subtraction/multiplication/division/constant multiple limit laws and the following standard limits:
$\displaystyle\lim_{n \to\infty}\left(n^{\frac{1}{n}}\right)=1$
$\displaystyle\lim_{n \to\infty}\left(c^{\frac{1}{n}}\right)=1$, where $c$ is a real number, and
$\displaystyle\lim_{n \to\infty}\left(1+\frac{a}{n}\right)^{n}=e^{a}$.
As well as those formed from the ratio of terms in the following hierarchy:
$$1<\log_e(n)<n^{p}<a^{n}<b^{n}<n!<n^{n}$$
Where $p>0$ and $1<a<b$
e.g. $\displaystyle\lim_{n \to\infty}\left(\frac{\log_e(n)}{n!}\right)=0$
At first I thought maybe I could express the limit in the form of the $e^{a}$ standard limit, but I couldn't seem to get rid of the square on $n$. I've also tried the sandwich theorem, it's obviously bounded above by $1$, but I just couldn't find any suitable lower bounds. I'd really appreciate some help, or even just a little hint, thanks!

$(1 - \frac1{n^2} )^{n^2} \to e^{-1}$ as $n \to \infty$.
Thus $(1 - \frac1{n^2} )^{n^2} \in [\frac12,1]$ as $n \to \infty$.
$\def\wi{\subseteq}$
Now $(1 - \frac1{n^2} )^n \in \left( (1 - \frac1{n^2} )^{n^2} \right)^\frac1n \wi [\frac12,1]^\frac1n \to {1}$ as $n \to \infty$.
Thus by the squeeze theorem $(1 - \frac1{n^2} )^n \to 1$ as $n \to \infty$.
