Matematické Fórum

Marian · 17. 07. 2014 17:10

Find the limit

Freedy · 17. 07. 2014 17:52

Limit:
$\lim_{n\to\infty }\frac{\pi }{2}\cdot\sqrt{n^2+n}$
$\lim_{n\to\infty }\frac{\pi }{2}\cdot\sqrt{n^2+n}=\frac{\pi }{2}\lim_{n\to\infty } n\sqrt{1+\frac{1}{n}}=\lim_{n\to\infty }\frac{\pi }{2}n$
násobky pi/2 jsou:
$\cos (0 +2k\pi )=0$
$\cos (\frac{\pi }{2}+2k\pi )=0$
$\cos (\pi +2k\pi )=-1$
$\cos (\frac{3\pi }{2} +2k\pi )=0$
Jelikož je tam na 4 tak se ze záporných hodnot stanou kladné. Limita neexistuje, bude se pohybovat v intervalu $\langle0;1\rangle$

Marian · 17. 07. 2014 19:30

↑ Freedy:

I'm afraid this is not a correct answer.

Freedy · 17. 07. 2014 20:59

Seriously?

If i take some high number, for example:
$\sqrt{n^2+n}=1000000$
so $n=\frac{-1+\sqrt{4000001}}{2}$
: $\cos ^4(500000\pi )=\cos ^4(250000(2\pi )) = 1$

I take some another high number, for example:
$\sqrt{n^2+n}=1000001$
so n = $n=\frac{-1+\sqrt{4000005}}{2}$
$\cos ^4(1000001\frac{\pi }{2})=\cos ^4(\frac{\pi}{2} +(2\pi )250000 )=\cos ^4(\frac{\pi}{2} ) = 0$
and this is different result... So there cant be exact value.

Xellos · 17. 07. 2014 21:48

It probably should've been mentioned that it's supposed to be a limit of a sequence, so $n \in \mathbb{N}$ , not $n \in \mathbb{R}$ , even though $n$ is almost never used for reals.

$\lim_{n \rightarrow \infty}{\sqrt{n^2+n}-n}=\lim_{n \rightarrow \infty}{\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}}=\lim_{n \rightarrow \infty}{\left(1+\sqrt{1+\frac{1}{n}}\right)^{-1}}=\frac{1}{2}$

Therefore, the argument of the cosine approaches $\frac{\pi}{2}n+\frac{\pi}{4}$ and the cosine alternates between $\frac{1}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$ , which gives the result: $\frac{1}{4}$ .

Marian · 18. 07. 2014 15:04

↑ Freedy:↑ Xellos:

The correct answer is, of course, 1/4.

However, from the title of my contribution 'Limit of a trigonometric sequence' it should be clear that the discussed limit is supposed to be the limit of a sequence.

check_drummer · 22. 07. 2014 23:05

↑ Xellos:
Hi, can you please explain the connection of your limit and the limit that should be computed? Thank you.

Brano · 23. 07. 2014 00:37 — Editoval Brano (23. 07. 2014 00:38)

↑ check_drummer:
maybe this will be more comprehensible (based on ↑ Xellos:)

$2\cos^2\left(\frac{\pi}{2}\sqrt{n^2+n}\right)-1=\cos(\pi\sqrt{n^2+n})=\cos(\pi[\sqrt{n^2+n}-n]+\pi n)=$
$=(-1)^n\cos\left(\pi\frac{n}{\sqrt{n^2+n}+n}\right)\to 0$
therefore
$\cos^4\left(\frac{\pi}{2}\sqrt{n^2+n}\right)\to\frac{1}{4}$

Xellos · 23. 07. 2014 02:41 — Editoval Xellos (23. 07. 2014 02:41)

↑ Marian:

Oh, the title. [insert facepalm here]

↑ check_drummer:

Well, since the argument of $\cos^4$ approaches $\frac{\pi}{2}n+\frac{\pi}{4}$ for large $n$ , and $\cos^4$ is a continuous function, then $\cos^4\left(\frac{\pi}{2}\sqrt{n^2+n}\right)$ will approach $\cos^4\left(\frac{\pi}{2}n+\frac{\pi}{4}\right)=\frac{1}{4}$ . It's kinda rough, but you should get the picture (yes, you should go and draw a graph of the situation :D).

Matematické Fórum

#1 17. 07. 2014 17:10

Limit of a trigonometric sequence

#2 17. 07. 2014 17:52

Re: Limit of a trigonometric sequence

#3 17. 07. 2014 19:30

Re: Limit of a trigonometric sequence

#4 17. 07. 2014 20:59

Re: Limit of a trigonometric sequence

#5 17. 07. 2014 21:48

Re: Limit of a trigonometric sequence

#6 18. 07. 2014 15:04

Re: Limit of a trigonometric sequence

#7 22. 07. 2014 23:05

Re: Limit of a trigonometric sequence

#8 23. 07. 2014 00:37 — Editoval Brano (23. 07. 2014 00:38)

Re: Limit of a trigonometric sequence

#9 23. 07. 2014 02:41 — Editoval Xellos (23. 07. 2014 02:41)

Re: Limit of a trigonometric sequence

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