- #1
izen
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Homework Statement
Homework Equations
The Attempt at a Solution
above picture !
Exam in 3 hours :( please help
jbunniii said:You can write
$$\begin{align}
1 - \cos\left(\frac{2}{n}\right) = 2\left[\frac{1}{2} - \frac{1}{2}\cos\left(\frac{2}{n}\right)\right]
&= 2\sin^2\left(\frac{1}{n}\right) \\
\end{align}$$
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$$
\frac{1 - \cos{\frac 1 n}}{\frac 1 n}$$
It didn't go anywhere. I was just suggesting how to rewrite the hard part. Now if we substitute back, we getizen said:where is 'n' gone?
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?
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.
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.
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.
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.
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.