4
$\begingroup$

What is $\lim_{n\rightarrow\infty}(1 + \frac{3}{n})^n$?

I'm a little confused on this limit. The $\frac{3}{n}$ part gets smaller as $n$ gets bigger, so it would really just come out to $1^n$, right?

$\endgroup$
3
  • 2
    $\begingroup$ Have you heard of the number $e$? $\endgroup$
    – Pedro Tamaroff
    Apr 20 '13 at 18:06
  • 2
    $\begingroup$ Just think of clever way to change it in the form $(1+\frac{1}{m})^{mc}$ where m is some function of $n$ and $c$ is a constant and use that. $\endgroup$
    – Alex Botev
    Apr 20 '13 at 18:07
  • $\begingroup$ You may also benefit from studying the answers to this question. $\endgroup$ Apr 20 '13 at 19:07
5
$\begingroup$

Hint: $$\lim_{n\to \infty} \left(1 + \frac 1n\right)^n = e\tag{1}$$

Try writing $\dfrac{3}{n} = \dfrac{1}{n/3}$, so the exponent $n = 3\cdot\dfrac n3$:

$$\lim_{n\to \infty} \left(1 + \frac 1{(n/3)}\right)^{3(n/3)}$$

Putting $m = \frac n3$ it looks very close to $(1)$:

$$\lim_{m\to \infty} \left(1 + \frac 1{m}\right)^{3m}$$

$$\lim_{n\to \infty} \left(1 + \frac 1{(n/3)}\right)^{3(n/3)}=\lim_{m\to \infty} \left(1 + \frac 1{m}\right)^{3m}= e^3$$

$\endgroup$
1
  • $\begingroup$ Thanks, Amzoti...for the compliment and for cluing me in on the formatting "issue" ;-) $\endgroup$
    – amWhy
    Apr 21 '13 at 0:44
3
$\begingroup$

First, look at why your observation fails. First, the exponent also goes to $\infty$, so you cannot really forget about it. It would be like saying $$1=\lim \frac{n}{n}=\lim n\frac 1n =\lim n\cdot 0=\lim 0=0$$ see?

For example, we agree that both $n^{-2}$ and $(\log n)^{-1}$ get small when $n\to\infty$; however

$$\lim \left(1+\frac{1}{\log n}\right)^n\to\infty $$

while

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

This has all a little to do with the fact that $$\lim_{n\to \infty} \left(1 + \frac 1n\right)^n = e$$

If you haven't heard of $e$ before, do a little reading about it. It turns out that $e\approx 2.718281828459045\dots$. In your case, the limit turns out to be related to $e$, since $$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{3}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{3}{n}} \right)^{\frac{n}{3}3}} = \cdots$$

$${\left[ {\mathop {\lim }\limits_{n \to \infty } {{\left( {1 + \frac{3}{n}} \right)}^{\frac{n}{3}}}} \right]^3} = {e^3}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.