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

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!

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.

• This is perfect! :) Apr 26 '16 at 7:06
• Thank you, this is brilliant! I knew there had to be a simple way! Apr 26 '16 at 7:09
• +1 You even only need that $\lim_{h\to 0}(1+h)^{1/h}$ exists at all, not that it $e$. Apr 26 '16 at 14:55
• @HagenvonEitzen But shouldn't you prove that $\lim_{h \to 0} (1-h)^{1/h} = \lim_{h\to 0} \frac 1{(1+h)^{1/h}}$? That does not seems so obvious to me
– Ant
Apr 26 '16 at 21:31
• @user21820: I agree with some part of your comment, namely that it is difficult to deal with real exponents compared to integer exponents. However if we know that $(1 + (1/n))^{n}$ tends to some limit $L$ then it is easy to show that $(1 - (1/n))^{n}$ tends to $1/L$. See math.stackexchange.com/a/1733139/72031 However, I show in my answer the current question does not need any information about $e$ or $\log$. Apr 27 '16 at 5:35

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 $1$.

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

• Why are you allowed to take the limits of a and b "separately" and put them together? Apr 26 '16 at 7:38
• Because you're not creating an indeterminate form. Apr 26 '16 at 9:19
• Because you can write $(1-1/n^2)^{n^2}=e^{-1}(1+d_n)$ where $d_n\to 0,$ so $|d_n|<1/2$ for all but finitely many $n.$ So $(1/2 e)^{1/n}<$ $(1-1/n^2)^n<$ $(3 /2e)^{1/n}$ for all but finitely many $n.$ Apr 26 '16 at 19:42
• @Nao: Because $\lim_{n\to\infty} (1-\frac{x}{n})^n =e^x$. Pushing the limit into the base and exponent will give nonsense instead. See my answer for how to rigorously push the analysis inside, by bounding instead and taking the limit later. Apr 27 '16 at 1:29
• Thanks for your answers! Makes sense now. Apr 27 '16 at 11:06

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

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

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

• The first line you wrote should approach to $e^{-1}$? Apr 26 '16 at 6:56
• @Solumilkyu: Yea I'm very careless today it seems.. Either way the idea is always the same; to move the limit past the exponent rigorously we simply bound instead of taking limit. Apr 26 '16 at 6:57

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

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.

• Nice! Don't know why this has no upvotes! You do need the convergence of the two separate limits, otherwise the whole reasoning breaks down. Aug 1 '16 at 14:41

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