Finding the Limit of a Sequence (No L'Hopital's Rule)

Solution 1:

Rewrite:

$\lim \limits_{n\to\infty} \left(1-{1\over n^2}\right)^n = \lim \limits_{n\to\infty} \left[\left(1-{1\over n}\right)\left(1+{1\over n}\right)\right]^n $

Distribute the powers over we have:

$ \lim \limits_{n\to\infty} \left(1-{1\over n}\right)^n\left(1+{1\over n}\right)^n= e^{-1}e=1 $ as desired.

Solution 2:

Let's use Bernoulli's Inequality to get $$1 - \frac{1}{n} \leq \left(1 - \frac{1}{n^{2}}\right)^{n} \leq 1\tag{1}$$ and then applying squeeze theorem as $n \to \infty$ we get the desired limit as $1$. Note that this particular limit is purely algebraic one and does not need any significant theorems / results dealing with $e$ or $\log$ function.

Bernoulli's Inequality: If $x \geq -1$ and $n$ is a non-negative integer then $$(1 + x)^{n} \geq 1 + nx\tag{2}$$ Here we have used this inequality with $x = -1/n^{2}$ to get the first inequality in $(1)$. The second inequality in $(1)$ is obvious as $0 \leq (1 - 1/n^{2}) \leq 1$ and hence any positive integral power of $(1 - 1/n^{2})$ does not exceed $1$.

Solution 3:

$$\lim_{n\to\infty} \left(1-{1\over n^2}\right)^n = \lim_{n\to\infty} \left[\left(1-{1\over n^2}\right)^{n^2}\right]^{1/n} $$ You have an exponential form $a^b$, where $a$ is approaching $e^{-1}$ [replace $n$ with $\sqrt n$ here] and $b$ is approaching $0$.

Solution 4:

One may use, as $x \to 0$, the classic Taylor series expansion $$ \begin{align} \log(1-x)&=-x+o\left(x\right), \end{align} $$ giving, as $n \to \infty$,

$$ n\log \left(1-\frac1{n^2}\right)=-\frac1{n}+o\left(\frac1n\right) $$

then observe that

$$ \lim_{n\to\infty} \left(1-{1\over n^2}\right)^n=\lim_{n\to\infty} e^{n\log \left(1-\frac1{n^2}\right)}=\lim_{n\to\infty}e^{-\frac1{n}+o\left(\frac1n\right)}=e^0=1. $$