Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$ Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$
I need to show that there exists $N \in \mathbb{N}: \forall n \geq N : \left|(1-\frac{1}{n^2})^n-1\right| \lt \epsilon$ $,\:\:\forall \epsilon \gt 0.$
Since $(1-\frac{1}{n^2})^n \leq 1$ $\forall n \in \mathbb{N}_+ \Rightarrow \left|(1-\frac{1}{n^2})^n-1\right| = 1 - (1-\frac{1}{n^2})^n$
Therefore I need to solve $1 - \left(1-\frac{1}{n^2}\right)^n < \epsilon$ for $n$.
Unfortunately I cannot solve this explicitly for n. The best I can do is:
$$\log(1-\epsilon) \lt n \log\left(1-\frac{1}{n^2}\right)$$
Is there a way to solve this for $n$ explicitly? How can I prove this if this is if it is not solvable for $n$?
 A: Hint: if $x\geqslant 0$ and $n$ is an integer, then $(1-x)^n\geqslant 1-nx$. 
This can be seen definigin $f(x)=(1-x)^n+nx$ and showing that the derivative is non-negative.
A: Hint:
\begin{equation}
\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{\Large n}=\lim_{n\to\infty}\left[\left(1-\frac{1}{n^2}\right)^{\Large n^2}\right]^{\Large\frac{1}{n}},
\end{equation}
where $\displaystyle\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{\Large n^2}=\frac {1}{e}$.
A: Consider the expansion of $(1-\frac{1}{n^2})^n.$
$$\begin{align}\left(1-\frac{1}{n^2}\right)^n &= 1 - n\dfrac{1}{n^2}+\dfrac{n(n-1)}{2}\dfrac{1}{n^4} - \cdots \\ &= 1-\dfrac{1}{n}+\dfrac{n-1}{2n^3}+\cdots\\&\geq1-\dfrac{1}{n}\end{align}$$
We can construct the inequality $$1-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right)^n \leq 1$$
Subtract $1$ and then change the order of inequality
$$-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right)^n -1 \leq 0$$
$$0 \leq 1- \left(1-\frac{1}{n^2}\right)^n \leq \dfrac{1}{n}$$
After this you can just choose $N = \left\lceil\dfrac{1}{\epsilon}\right\rceil.$
