# Evaluate limit using series

1. Mar 11, 2009

### wilcofan3

1. The problem statement, all variables and given/known data

Evaluate $$\lim_{n->\infty} \frac {cos(n+1)} {ln n}$$

2. Relevant equations

$$cos (x+1) = 1 - \frac {(x+1)^2} {2!} + \frac {(x+1)^4} {4!} - ...$$

$$ln x = (x-1) - \frac {(x-1)^2} {2} + \frac {(x-1)^3} {3} - \frac {(x-1)^4} {4} + ...$$

3. The attempt at a solution

Using those formulas, and then canceling, here is my pitiful attempt:

$$\lim_{n->\infty} \frac {1 - \frac {1} {n!}} {(n-1)}$$

2. Mar 11, 2009

### Staff: Mentor

Can't you just take the limit directly? The numerator is always between -1 and 1, while the denominator grows large without bound.

3. Mar 11, 2009

### wilcofan3

I just didn't think I would simply be evaluating a limit in the section of Calculus that I'm in, so I assumed there was more to it involving series or something.

Any ideas?

4. Mar 12, 2009

### Staff: Mentor

Does the problem say you have to use series representations? If not, the approach I gave is much simpler.

5. Mar 12, 2009

### wilcofan3

The problem is from a worksheet that appears in the section of the book about series, but all the problem says is "Evaluate the limit." The other problems on the sheet are about series, and that's why I thought I had to do it this way.

6. Mar 12, 2009

### Staff: Mentor

Limits often show up when you're asked to determine whether a given series converges or diverges, so maybe this limit will show up in a later problem on this sheet.

7. Mar 12, 2009

### wilcofan3

Thanks.

So, just to make sure, since the numerator is between the interval -1<x<1 and the denominator goes off to $$\infty$$, this would mean the limit is going to 0, correct?

Is there anything I need to specifically show other than those aspects?

8. Mar 12, 2009

### Staff: Mentor

Yes, correct.
You can use what some textbooks call the "Squeeze" or "Squeeze Play" theorem.

You can bound your limit expression like so:
-1/(ln n) <= cos(n + 1)/(ln n) <= 1/(ln n)

The limit, as n approaches infinity of the expression is 0, and the limit of the expression on the right is also 0, which means that the expression in the middle has the same limit.

9. Mar 12, 2009

### wilcofan3

Thanks! Hopefully this is all that's required.

10. Mar 12, 2009

### Staff: Mentor

If you're going to assume anything, assume that you should work the problem in the simplest way you can get away with. By "get away with," I don't include any technique that contravenes explicit requirements in the problem.