# Limits again

1. Jan 18, 2008

### Firepanda

http://img186.imageshack.us/img186/4594/howtohe5.jpg

I know the limits of each are 1 and 3 respectively, and I did work them out by myself, but the question is 8 marks so I'm sure I didn't have enough correct notation, as I had like a line for each and I used more words than math.

What would the correct notation be for questions like these? My book has no examples for such.

2. Jan 18, 2008

### unplebeian

What do you mean by enough correct notation? 'Therefore' signs and 'since' signs?

How did you use more words than math?

I am not sure what you mean by correct notations, but if you want a hint how to start solving these problems apply logs on both sides. By the way have you stuided indeterminate forms?

3. Jan 18, 2008

### Firepanda

Well for the first problem as n tend to infinity, 3 becomes negligible, therefore from the definition of the limit for n^1/n in the question, it also tends to 1. But what is the notation for this?

Its pretty much the same for the other aswell.

Thanks

4. Jan 18, 2008

### unplebeian

I am still unclear about what you mean. After you know that the limit is an indeterminate form you can apply L'Hopitals rule and then finally figure out that the limit is 1. You need to SIMPLIFY the expression inorder to figure out the answer. Simply 'estimating' by n^2/n etc, it isn't going to do it.

5. Jan 18, 2008

### Firepanda

Sorry how do I use L'Hopitals here? :P

I only used it in forms of f(x)/g(x).

6. Jan 18, 2008

### unplebeian

7. Jan 18, 2008

### dynamicsolo

You need to put each of these functions through a sequence of transformations to get it into a form suitable to apply l'Hopital's Rule.

First, take the natural log of the function (important: remember that you did this!). For the first problem, you will now have (1/n) · [ln{(n^2)+3}]. This is an indeterminate product in the limit as n approaches infinity. So we now rewrite it as an indeterminate ratio [ln{(n^2)+3}] / n , which gives inf./inf. You can now apply l'Hopital easily to this. The limit you get is the natural logarithm of the limit of your original function, so that limit is e^[your result].

The second one will be a bit more work, but will yield to the same method. (However, if you look at what is important in the argument of the radical as n becomes large, you can probably anticipate the result...)