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

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

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