# Infinite limits on a sequence

1. Apr 21, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Well, Hi again. Here I got some limits to infinity that I don't know how to solve. The statement just ask me to calculate those limits, if exists, for the next sequences.

$$\displaystyle\lim_{n \to{+}\infty}{(\sqrt[n]{n+1})}$$

$$\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{\sin(n)}{n+1}}$$

$$\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{1-\cos(5n)}{n^2}}$$

I don't pretend that someone solve all of these for me, but I need some orientation on which way to look. I'll be thanked for any help on any of those sequences.

Edit: Deleted the ones that I could solve. Someone gave me some tips, and I could solve some of the limits that I have posted before. As there wasn't any responses to the topic I thought that it was unnecessary to repost giving the advertisement.

Last edited: Apr 21, 2010
2. Apr 21, 2010

### LCKurtz

Hints: Think about logarithms on the first and think about how big sines and cosines can be on the others.

3. Apr 21, 2010

### Telemachus

I resolved the trigonometric ones. Thanks for the hints, I'll keep working on the first.

I established on the last two that sine and cosine are between 1 and -1, so both tend to zero.

On the first, is this valid? $$(\sqrt[n]{n+1})=\displaystyle\frac{log (n+1)}{n}\$$

Last edited: Apr 21, 2010
4. Apr 21, 2010

### LCKurtz

No, those two quantities aren't equal are they? But you could call

$$y = \left({n+1}\right)^{\frac 1 n}$$

and then say

$$\ln(y) = \frac{log (n+1)}{n}$$

and work with ln(y).

5. Apr 21, 2010

### Telemachus

Thanks, I'll remember that, because it looks like the legs of a woman (y) :D

How did you realize that applying the ln to the expression gives the same function?

ln(y) tends to infinite really fast. How do I get the relation between what happens with ln(y) and y?

Last edited: Apr 21, 2010