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

[izen]

1. Homework Statement

2. Homework Equations



3. The Attempt at a Solution

above picture !!

Exam in 3 hours :( please help

 

[jbunniii]

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]

$$\frac 1 {\frac 1 n}$$doesn't go to 0 as $n\to\infty$.

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

 

[izen]

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?

 

[izen]

$$\frac 1 {\frac 1 n}$$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]

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?

 

[izen]

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