# Homework Help: An analysis question

1. Oct 21, 2012

### Artusartos

In this link,

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw7sum06.pdf [Broken]

for question 20.18, I wasn't sure about how the solution proved that the limited actually exsited. It does say what the limit is, but where does it prove that the limit exists?

Thanks in advance

Last edited by a moderator: May 6, 2017
2. Oct 21, 2012

### bossman27

The two steps here (proving existence and finding a value) are pretty intertwined. The only way you could have an indeterminate form here is if you had 0 in the denominator, and if you show that you can write the equation such that you don't have any division by 0, you've shown that the limit exists.

There may be a more elegant proof to point to, but in this case I think it's irrelevant; as long as you've shown that the denominator, numerator, and fraction are defined, you've proved existence.

3. Oct 21, 2012

### Artusartos

So to prove the existence of a limit, is it enough to show that the function is (or can be written in a different way that is) defined at the limit point?

4. Oct 21, 2012

### bossman27

As I said, there may be a more elegant way to show this proof, but there certainly is no question as to existence once you've done that. It's especially obvious when we're talking about the limit of x as it tends toward 0, rather than infinity.

Generally speaking, showing existence of a limit of a sequence (or series) involves seeing whether indeterminate forms can be reduced to determinate ones. In the original equation, we had a denominator of 0, which means the fraction is in an indeterminate form. If you can manipulate the equation such that you remove all 0 denominators, infinities, etc... you have shown that the limit is determinate. And a limit is determinate if and only if it exists.

5. Oct 21, 2012

### Artusartos

Alright, thanks a lot...

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook