# An analysis question

1. Oct 21, 2012

### Artusartos

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?

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...