Finding a basic limit

Question

Determine the following limit

$$\lim_{n\to\infty} \cos{\left(\frac{n}{2^{n}}\right)}$$

I'm not really sure how to start here. We can write this as $$\cos{\left(\lim_{n\to\infty}\frac{n}{2^{n}}\right)}$$ then we must determine $$\lim_{n\to\infty}\frac{n}{2^{n}}$$ evidently this is zero however how would we actually determine this?

Basic theorem: if ${a_n}$ is a converging sequence and $f(x)$ is a continuous function, $\lim_{n\to +\infty}f(a_n) = f\left(\lim_{n\to+\infty} a_n\right).$ — Jack D'Aurizio, Dec 28 '14 at 16:58

score 3 · Answer 1 · answered Dec 28 '14 at 17:23

3

The answer is $1$ because:

The function $\cos x$ is continous on $\mathbb{R}$ and at $x = 0$.
$\displaystyle \lim_{x\to 0} \cos x = \cos 0 = 1$.
$\displaystyle \lim_{n \to \infty} \dfrac{n}{2^n} = 0$ ( by L'hospitale rule )

answered Dec 28 '14 at 17:23

DeepSea

77,651

1

A function $f:\mathbb N\longrightarrow \mathbb R$ is not derivable ! – idm Dec 28 '14 at 17:46
@idm yes but if $f$ converges to $0$ on $\mathbb{R}$ then it also converges restricted to $\mathbb{N}$ – obataku Dec 28 '14 at 17:54
of course it does (that's why I didn't downvote), but it's tricky ! – idm Dec 28 '14 at 18:29

obataku · Accepted Answer · 2014-12-28T17:50:42.793

1

Do you know that $\lim\limits_{x\to a}f(g(x))=\lim\limits_{u\to b}f(u)$ where $b=\lim\limits_{x\to a}g(x)$ for sufficiently well-behaved $f,g$?

To prove $n/2^n\to0$ just recall that $2^n>n^2$ for $n>5$ so $0<\dfrac{n}{2^n}<\dfrac{n}{n^2}=\dfrac1n$ for sufficiently large $n$ and by the squeeze theorem from $0,1/n\to0$ it follows $n/2^n\to0$.

Of course, this relies on knowing $1/n\to0$. Do you have that result? If not it might be best to take a more direct epsilon-delta approach.

edited Dec 28 '14 at 17:50

answered Dec 28 '14 at 16:57

obataku

5,571

Yes I understand this thanks. I have made an edit to my post with further information. – user2850514 Dec 28 '14 at 17:02
I've updated my post. – obataku Dec 28 '14 at 17:51

score 1 · Answer 3 · answered Dec 28 '14 at 17:18

1

Since $\sum\limits_{n=1}^{\infty}\frac{n}{2^n}$ converges (use for example ratio test), we have $\lim_{n\to\infty}\frac{n}{2^n} = 0$.

answered Dec 28 '14 at 17:18

ChocolateBar

971

k170 · Answer 4 · 2014-12-28T17:32:41.530

1

Here are the steps $$\lim_{n\to\infty} \cos{\left(\frac{n}{2^{n}}\right)}= \cos{\left( \lim_{n\to\infty} \frac{n}{2^{n}}\right)} $$ $$=\cos\left(\lim_{n\to\infty}\frac{\frac{d}{dn}n}{\frac{d}{dn}2^{n}}\right)=\cos\left(\lim_{n\to\infty}\frac{1}{2^{n}\ln 2}\right) $$ $$=\cos(0)=1$$

edited Dec 28 '14 at 17:32

answered Dec 28 '14 at 17:22

k170

9,045

1

A function $f:\mathbb N\longrightarrow \mathbb R$ is not derivable ! – idm Dec 28 '14 at 17:46
1

@idm you mean differentiable? – Karl Dec 28 '14 at 18:03

score 0 · Answer 5 · answered Dec 28 '14 at 17:44

If $x_n=\frac{n}{2^n}$,

$$\left|\frac{x_{n+1}}{x_n}\right|=\left|\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}\right|=\left|\frac{(n+1)2^n}{n 2^{n+1}}\right|=\left|\frac{n+1}{n}\cdot \frac{1}{2}\right|\underset{n\to\infty }{\longrightarrow} \frac{1}{2}<1,$$ and thus $$\lim_{n\to\infty }\frac{n}{2^n}=0.$$

Finding a basic limit

5 Answers5