# Homework Help: What ? Limit of 0/y as y tends to 0 is 0?

1. Sep 21, 2006

### precondition

What!!?? Limit of 0/y as y tends to 0 is 0???

How come? Limit of 0/y as y tends to 0 is 0??
I thought 0/y as y tends to infinity is 0 and above case causes problems due to limiting process...
what's epsilon delta argument for this??

2. Sep 21, 2006

### StatusX

0/y=0 for all y not equal to zero. The argument's pretty clear from here.

3. Sep 21, 2006

### quasar987

The epsilon-delta argument is that given e>0, for 0<|y|<d=e,

|0/y|=0/|y|=0<e

In words, 0/y is 0 for any y other than 0. and the limit of 0 as y approches anything is 0.

4. Sep 21, 2006

### precondition

:O
0/y=0 for all y not equal to zero. YES
The argument's pretty clear from here. ????????????????????????

5. Sep 21, 2006

### moose

If y is anything except for 0, then it will equal 0. So if y approaches 0 without equaling 0, then it will be 0....

6. Sep 21, 2006

### StatusX

For any epslion>0, you can take delta to be whatever you want, since f(y)=0/y is within epsilon for all y, namely because it is equal to it.

The intuitive idea of a limit is that as you get closer and closer to a certain point (y=0 here, n=infinity (sort of) in the case of limits of infinite sequences), the values of the expression you're taking the limit of get closer and closer to some limiting value, ie, the limit. So if all the terms are equal, there is obviously only one thing this limit can be.

Last edited: Sep 21, 2006
7. Sep 21, 2006

### precondition

yes... but I always thought when we think about limits, when we say as y tends to 0 then we can't think of y as 0 but also I thought we can't think of y as not 0?? ummm... then you may think what do I think of y as? good question....-_-;;

8. Sep 21, 2006

### StatusX

I don't know what you're saying. By the way, for these kinds of general questions about math, it's probably better to ask them in the general math forum down below.

9. Sep 21, 2006

### quasar987

We have to think of y as not zero. The limit definition says that the limit of f(y) as y goes to 0 is L if no matter how small a number e we choose, there exists an interval around y=0 (but excluding 0) such that for all y in that interval, the distance from f(y) to L is smaller than e.

10. Sep 21, 2006

### precondition

oh ok.... so let me try
so, consider f(y)=0/y. By definition of limit states that for all epsilon>0 there exists delta such that lf(y)-0l<e for all ly-0l<d so we have l0/y-0l<e for lyl<d so let d=e then, l0/yl=0/lyl>0/d??? umm.. at this step, don't we have ">" which makes the argument fail?

11. Sep 21, 2006

### quasar987

This is not how you must reason. Given an e>0, you're looking for a number d(e) such that when |y|<d, then the inegality |0/y-0|=0<e is verified. But the inegality 0<e is verified independantly of what y is, since e>0 is how we "defined" e in the begining. So you can choose d to be anything.

12. Sep 21, 2006

### precondition

ok 0/lyl is already 0 so any epsilon covers it so any delta will do. Hmm... still i have this feeling that I didin't flush the toilet...

13. Sep 21, 2006

### istevenson

man,Limit of 0/y as y tends to 0 is 0
it is very clear.

14. Sep 21, 2006

### ToxicBug

Just do l'Hopital's: 0/y => 0/1 = 0... n e more questions?

15. Sep 21, 2006

### precondition

istevenson,
does your mom tell you 'man it's all obviously clear' everytime you ask a question? If not, don't reply if you can't provide an 'explanation'

16. Sep 21, 2006

### StatusX

Maybe you can think about it this way: if two functions f(x) and g(x) are equal for all x$\neq$a, then:

$\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} g(x)$

Do you understand why this is true? If so, just take f(y)=0/y, g(y)=0, and a=0.

17. Sep 21, 2006

### precondition

statusx,
thank you i think epsilon delta argument makes sense.

18. Sep 21, 2006

### Dragonfall

My mom tells me 'man it's all obviously clear' every time I ask a question.

If you plot 0/y =f(y), you will see that it's defined everywhere (as 0) everywhere except at y=0. Now adding a limit at y=0 should make f(y) continuous, if such a limit exists. If you add anything except f(0)=0, will the function be continuous?

Intuitive arguments are better than delta-epsilon ones.

19. Sep 22, 2006

### istevenson

precondition,do you know what is the difference between the following two expressions:
$$\lim_{y\rightarrow0} \frac{0}{y}$$

$$\lim_{y\rightarrow0}\frac{f(y)}{y}$$

supposing when $$y\rightarrow0$$ f(y) $$\rightarrow0$$

If you do,i will tell you that your problem is really really obviously clear.

Last edited: Sep 22, 2006
20. Sep 22, 2006

### HallsofIvy

No, that's not quite the definition of limit
Correct is "for all epsilon>0 there exists delta such that lf(y)-0l<e for all
0< ly-0l<d so we have l0/y-0l<e". Do you see the difference? It's that
"0< |y-0|" part. What happens when y= 0 is irrelevant.

Also, your "for lyl<d so let d=e then, l0/yl=0/lyl>0/d" has some bad algebra. |y|< d does NOT imply |a/|y||> a/d. It depends on whether a is positive, negative, or 0. Here, since a= 0, what is correct is that
|0/|y||= 0/d.

You don't need to take d= e. Since, for y not 0, 0/y= 0, you have
|0/y-0|= 0< e for any d.