• Support PF! Buy your school textbooks, materials and every day products Here!

Limit of sequence lim n (1-cos(2/n))

  • Thread starter izen
  • Start date
51
0
1. Homework Statement

Untitled.jpg


2. Homework Equations



3. The Attempt at a Solution

above picture !!

Exam in 3 hours :( please help
 

jbunniii

Science Advisor
Homework Helper
Insights Author
Gold Member
3,386
179
You can write
$$\begin{align}
1 - \cos\left(\frac{2}{n}\right) = 2\left[\frac{1}{2} - \frac{1}{2}\cos\left(\frac{2}{n}\right)\right]
&= 2\sin^2\left(\frac{1}{n}\right) \\
\end{align}$$
 

LCKurtz

Science Advisor
Homework Helper
Insights Author
Gold Member
9,488
723
$$\frac 1 {\frac 1 n}$$doesn't go to 0 as ##n\to\infty##.

Another method is to use L'Hospital's rule on$$
\frac{1 - \cos{\frac 1 n}}{\frac 1 n}$$
 
51
0
You can write
$$\begin{align}
1 - \cos\left(\frac{2}{n}\right) = 2\left[\frac{1}{2} - \frac{1}{2}\cos\left(\frac{2}{n}\right)\right]
&= 2\sin^2\left(\frac{1}{n}\right) \\
\end{align}$$
where is 'n' gone?
 
51
0
$$\frac 1 {\frac 1 n}$$doesn't go to 0 as ##n\to\infty##.

Another method is to use L'Hospital's rule on$$
\frac{1 - \cos{\frac 1 n}}{\frac 1 n}$$
thanks LCKurtz
 

jbunniii

Science Advisor
Homework Helper
Insights Author
Gold Member
3,386
179
where is 'n' gone?
It didn't go anywhere. I was just suggesting how to rewrite the hard part. Now if we substitute back, we get
$$n\left[1 - \cos\left(\frac{2}{n}\right)\right] = 2n \sin^2\left(\frac{1}{n}\right)$$
Can you see what to do now?
 
51
0
It didn't go anywhere. I was just suggesting how to rewrite the hard part. Now if we substitute back, we get
$$n\left[1 - \cos\left(\frac{2}{n}\right)\right] = 2n \sin^2\left(\frac{1}{n}\right)$$
Can you see what to do now?

Thanks jbunniii
 

Related Threads for: Limit of sequence lim n (1-cos(2/n))

  • Last Post
Replies
10
Views
5K
  • Last Post
Replies
9
Views
5K
Replies
3
Views
621
Replies
8
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
4K
Replies
4
Views
942
Top