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

Question

Okay, I feel almost silly for asking this, but I've been on it for a good hour and a half. I need to find: $$\lim_{n \to\infty}\left(1-\frac{1}{n^{2}}\right)^{n}$$ But I just can't seem to figure it out. I know its pretty easy using L'Hopital's rule, and I can "see", that it's going to be $1$, but apparently it's possible to compute using only the results I've proved so far. That is, the sandwich theorem, the addition/subtraction/multiplication/division/constant multiple limit laws and the following standard limits: $\displaystyle\lim_{n \to\infty}\left(n^{\frac{1}{n}}\right)=1$ $\displaystyle\lim_{n \to\infty}\left(c^{\frac{1}{n}}\right)=1$, where $c$ is a real number, and $\displaystyle\lim_{n \to\infty}\left(1+\frac{a}{n}\right)^{n}=e^{a}$. As well as those formed from the ratio of terms in the following hierarchy: $$1<\log_e(n)<n^{p}<a^{n}<b^{n}<n!<n^{n}$$ Where $p>0$ and $1<a<b$

e.g. $\displaystyle\lim_{n \to\infty}\left(\frac{\log_e(n)}{n!}\right)=0$

At first I thought maybe I could express the limit in the form of the $e^{a}$ standard limit, but I couldn't seem to get rid of the square on $n$. I've also tried the sandwich theorem, it's obviously bounded above by $1$, but I just couldn't find any suitable lower bounds. I'd really appreciate some help, or even just a little hint, thanks!

Answer 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.

Answer 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.

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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 $1$.

Answer 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 $0$.

Answer 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

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

Answer 5

$(1 - \frac1{n^2} )^{n^2} \to e^{-1}$ as $n \to \infty$.

Thus $(1 - \frac1{n^2} )^{n^2} \in [\frac12,1]$ as $n \to \infty$. $\def\wi{\subseteq}$

Now $(1 - \frac1{n^2} )^n \in \left( (1 - \frac1{n^2} )^{n^2} \right)^\frac1n \wi [\frac12,1]^\frac1n \to {1}$ as $n \to \infty$.

Thus by the squeeze theorem $(1 - \frac1{n^2} )^n \to 1$ as $n \to \infty$.

Answer 6

Fix $x>0$, and take $N$ large enough such that $\frac{1}{N}<a$. Then, for every $n\geq N$ you have $$\left(1-\frac{a}{n}\right)\leq \left(1-\frac{1}{n^2}\right)\leq \left(1+\frac{a}{n}\right)$$ (The first inequality is true because $\frac{1}{n}<a$. The second inequality is true because the left hand side is less than 1, while the right hand side is greater than 1).

Thus, we have

\begin{align*} \left(1-\frac{a}{n}\right)^n&\leq \left(1-\frac{1}{n^2}\right)^n\leq \left(1+\frac{a}{n}\right)^n\\ \lim_{n\to\infty}\left(1-\frac{a}{n}\right)^n&\leq \lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n\leq \lim_{n\to\infty} \left(1+\frac{a}{n}\right)^n\\ e^{-a}&\leq \lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n\leq e^a \end{align*}

Now, since this is true for every $a>0$, we can take the limit $a\to 0$, obtaining

$$1=\lim_{a\to 0}e^{-a}\leq \lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n\leq \lim_{a\to 0}e^a=1.$$

Remark: In the last line, notice that $\displaystyle{\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n}$ does not depend on $a$, so we have $$\displaystyle{\lim_{a\to 0}\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n=\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^n}.$$

Answer 7

Here's an argument that doesn't require any special concepts like Taylor Series or the limit definition of $e$:

$$\begin{align} \lim_{n\to\infty} \left(1-\frac1{n^2}\right)^n &= \lim_{n\to\infty} \left(1-\frac1n\right)^n\left(1+\frac1n\right)^n\\ &= \lim_{n\to\infty} \left(\frac{n-1}n\right)^n\times \lim_{m\to\infty}\left(\frac{m+1}m\right)^m\\ &= \lim_{n\to\infty} \left(\frac{n-1}n\right)^n\times \lim_{p\to\infty}\left(\frac{p}{p-1}\right)^{p-1}\\ &= \lim_{n\to\infty} \left(\frac{n-1}n\right)^n\left(\frac{n}{n-1}\right)^{n-1}\\ &= \lim_{n\to\infty} \left(\frac{n-1}n\right)\\ &= 1 \end{align}$$ Here, I've split the limit of the product into the product of limits, temporarily, using $m$ for the new variable. Then, I've let $m=p-1$, and then recombined the product.

The only thing this approach is conditional on is the convergence of the two separate limits, and may not even require that much.

Answer 8

This is a very useful and simple method: $$Since \lim_{n \to\infty}\left(1-\frac{1}{n^{2}}\right)^{n}$$ And $$ n {\to\infty},$$ We can infer that $$ \frac {1}{n^2} {\to 0} $$ Since our whole argument of limits is based on approximation, we may use the (very) common approximation used in physics and mathematics, where $$({1-a})^x =({1-xa}) $$ Provided a is an extremely insignificant value. Therefore, $$\lim_{n \to\infty}\left(1-\frac{1}{n^{2}}\right)^{n}= \lim_{n \to\infty}\left(1-\frac{n}{n^{2}}\right)= \lim_{n \to\infty}\left(1-\frac{1}{n}\right).$$ $$Substituting \ n \ as \ \infty, \frac {1}{n} =0$$. Therefore, 1 is the required answer