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!