$\lim \limits_{n \to \infty} \left(1-\frac1{n^2}\right)^{n}$ equals

Question

The limit $\lim \limits_{n \to \infty} \bigg(1-\dfrac{1}{n^2}\bigg)^{n}$ equals?

$(A) e^{-1}$

$(B) e^{-\frac{1}{2}}$

$(C) e^{-2}$

$(D) 1$

Using the expansion $(1+x)^{n}=1+nx+ \dfrac{n(n-1)x^{2}}{2!} ...$
I am having large degree of $n$ in denominator therefor i am left with $1$ after applying the limit. Did i do it right ? If i did it right is there any alternative method to solve this problem like a quick version?

Answer 1

Yes, $\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^n=\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty}\left(\left(1+\frac{1}{-n}\right)^{-n}\right)^{-1}\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e^{-1}\cdot e=1$

Answer 2

Hint:

$$\lim_{n\to\infty}\left(1-\dfrac1{n^2}\right)^n=\left(\lim_{n\to\infty}\left(1+\dfrac1{-n^2}\right)^{-n^2}\right)^{\lim_{n\to\infty}-1/n}$$

Set $x=1$ in About $\lim \left(1+\frac {x}{n}\right)^n$

Answer 3

$1-\frac{1}{n^2} \to 1$ and $n \to \infty$, so: $$\lim_{n \to \infty}\left(1-\frac{1}{n^2}\right)^n=\exp\left(\lim_{n \to \infty}\left(1-\frac{1}{n^2}-1\right)n\right)=\exp\left(\lim_{n \to \infty}\left(-\frac{1}{n^2}\right)n\right)=\exp\left(\lim_{n \to \infty}-\frac{1}{n}\right)=\exp(-0)=\exp(0)=1$$

Answer 4

By Bernoulli: $ \left(1-\dfrac1{n^2}\right)^n \ge 1-\frac{1}{n}$, hence

$$1-\frac{1}{n} \le \left(1-\dfrac1{n^2}\right)^n \le 1.$$

Conclusion ?