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

In summary, the conversation discusses finding the limit as n approaches infinity, using L'Hospital's rule and rewriting the equation to make it easier to solve. The participants suggest using trigonometric identities and substitution to simplify the equation and solve for the limit.
  • #1
izen
51
0

Homework Statement



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Homework Equations





The Attempt at a Solution



above picture !

Exam in 3 hours :( please help
 
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  • #2
You can write
$$\begin{align}
1 - \cos\left(\frac{2}{n}\right) = 2\left[\frac{1}{2} - \frac{1}{2}\cos\left(\frac{2}{n}\right)\right]
&= 2\sin^2\left(\frac{1}{n}\right) \\
\end{align}$$
 
  • #3
$$\frac 1 {\frac 1 n}$$doesn't go to 0 as ##n\to\infty##.

Another method is to use L'Hospital's rule on$$
\frac{1 - \cos{\frac 1 n}}{\frac 1 n}$$
 
  • #4
jbunniii said:
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?
 
  • #5
LCKurtz said:
$$\frac 1 {\frac 1 n}$$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
 
  • #6
izen said:
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?
 
  • #7
jbunniii said:
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
 

1. What is the limit of the sequence lim n (1-cos(2/n)) as n approaches infinity?

The limit of the sequence as n approaches infinity is equal to 2. This can be verified by using the L'Hospital's rule or by using the trigonometric identity lim x→0 (1-cosx)/x = 0.

2. How do you prove that the limit of the sequence lim n (1-cos(2/n)) is equal to 2?

To prove that the limit is equal to 2, we can use the squeeze theorem. By rewriting the sequence as 2(1-cos(2/n))/n and using the fact that cos(2/n) is always between -1 and 1, we can show that the limit is bounded between 2 and 2. Therefore, the limit is 2.

3. Can you find the limit of the sequence lim n (1-cos(2/n)) without using L'Hospital's rule?

Yes, we can use the trigonometric identity lim x→0 (1-cosx)/x = 0 to find the limit. By substituting 2/n for x, we get the limit of 2, which is the same result as using L'Hospital's rule.

4. What is the significance of the limit of the sequence lim n (1-cos(2/n))?

The limit of this sequence has connections to the concept of limits and trigonometric identities. It is also used in calculus to find the derivatives of functions involving trigonometric functions.

5. Can the limit of the sequence lim n (1-cos(2/n)) be generalized to other sequences?

Yes, the limit of this sequence can be generalized to other sequences involving trigonometric functions. For example, lim n (1-cos(1/n)) would also have a limit of 2. However, the limit may be different if the sequence involves other functions.

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