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!

Fix $x>0$, and take $N$ large enough such that $\frac{1}{N}<a$. Then, for every $n\geq N$ you have $$\left(1-\frac{a}{n}\right)\leq \left(1-\frac{1}{n^2}\right)\leq \left(1+\frac{a}{n}\right)$$
(The first inequality is true because $\frac{1}{n}<a$. The second inequality is true because the left hand side is less than 1, while the right hand side is greater than 1).
Thus, we have
\begin{align*}
\left(1-\frac{a}{n}\right)^n&\leq \left(1-\frac{1}{n^2}\right)^n\leq \left(1+\frac{a}{n}\right)^n\\
\lim_{n\to\infty}\left(1-\frac{a}{n}\right)^n&\leq \lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n\leq \lim_{n\to\infty} \left(1+\frac{a}{n}\right)^n\\
e^{-a}&\leq \lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n\leq e^a
\end{align*}
Now, since this is true for every $a>0$, we can take the limit $a\to 0$, obtaining
$$1=\lim_{a\to 0}e^{-a}\leq \lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n\leq \lim_{a\to 0}e^a=1.$$
Remark: In the last line, notice that $\displaystyle{\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n}$ does not depend on $a$, so we have $$\displaystyle{\lim_{a\to 0}\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n=\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n}.$$
