The limit $ \lim_{n \to \infty} (1- \frac{1}{n^{2}} )^{n} $ The limit 
$ \lim_{n \to \infty}  (1- \frac{1}{n^{2}} )^{n} $
So how to do this problem? We cant transform into $(1+\frac1x)^x $ @Macavity Sir please help?
How to approach? 
 A: $$\left(1-\frac1{n^2}\right)^n=\left[\left(1-\frac1{n^2}\right)^{-n^2}\right]^{-1/n}$$
Now for the inner part, $\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$
Now check the limit of the exponent 
A: What about a proof, which does not even require any knowledge about the convergence of $\left(1+\frac{1}{n}\right)^n$:
We have the sandwhich:
$$1 \geq \left(1-\frac{1}{n^2}\right)^n \geq 1-\frac{1}{n}$$
, where the latter inequality follows from the well known (Maybe the first inequality taught in any first calculus course) Bernoulli's inequality.
A proof using $\left(1+\frac{1}{n}\right)^n \to e$ (and even $\left(1-\frac{1}{n}\right)^n \to e^{-1}$) is an overkill imho, since it uses the mathematical input, that those sequences converge (which needs some effort).
A: hint: $\left(1-\dfrac{1}{n^2}\right) = \left(1-\dfrac{1}{n}\right)\left(1+\dfrac{1}{n}\right)$
A: Hint: (another possibility) write $$\left(1-\frac{1}{n^2}\right)^n = e^{n\ln\left(1-\frac{1}{n^2}\right)}$$
and use the Taylor series expansion of $\ln(1-x)$: $\ln(1-x)=-x+o(x)$.
A: You can do it like that :
$(1-\frac{1}{n^2})^n = e^{n \ln(1-\frac{1}{n^2}) } = e^{n (-\frac{1}{n^2}+O(\frac{1}{n^4})) }$
$= e^{-\frac{1}{n}+O(\frac{1}{n^3})} \to e^0 = 1$
A: so here $ -n^{2}{\to \ -\infty}$ . so the formula does not hold..$(1+\frac1n)^n $tends to e when $n{\to \infty}$  not $-\infty$..
