$\lim_{n\to \infty}\left(1 - \frac {1}{n^2}\right)^n =?$

Question

Can you give any idea regarding the evaluation of the following limit?

$\lim_{n\to \infty}\left(1 - \frac {1}{n^2}\right)^n$

We know that $\lim_{n\to \infty}\left(1 - \frac {1}{n}\right)^n = e^{-1}$, but how do I use that here?

Answer 1

Here is a hint:

$\left(1 - \dfrac{1}{n^2}\right) = \left(1 - \dfrac{1}{n}\right)\left(1 + \dfrac{1}{n}\right)$

Also use the fact that $(ab)^n = a^n b^n$.

Answer 2

Note that $$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n^2}=\frac{1}{e}.$$ It follows that for large enough $n$, we have $$\frac{1}{10}\lt\left(1-\frac{1}{n^2}\right)^{n^2}\lt 1.$$ Thus if $a_n$ is our expression, then for large enough $n$ we have $$\left(\frac{1}{10}\right)^{1/n}\lt a_n\lt 1.$$ Now Squeeze.

Answer 3

Hint: We can rewrite it as $\lim \limits_{n \to \infty} \left( (1 - \frac{1}{n^2} )^{n^2} \right)^{1/n}$

Answer 4

General approach for ${(1^{\pm})}^{\infty}$ :

$$f^g=e^{g\ln f}=e^{g \ln(1+f-1)}=e^{g(f-1)}$$

where I have used $\ln(1+x)=x-x^2/2...$ I have ignored other small terms.

Here, $f=1-\frac 1 {n^2}$ and $g=n$