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

#### izen

1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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#### jbunniii

Homework Helper
Gold Member
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

Homework Helper
Gold Member
$$\frac 1 {\frac 1 n}$$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

Homework Helper
Gold Member
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