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

[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??

 

[StatusX]

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

 

[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.

 

[precondition]

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

 

[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...

 

[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.

 

[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...-_-;;

 

[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.

 

[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.

 

[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?

 

[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.

 

[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...

 

[istevenson]

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

 

[ToxicBug]

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

 

[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'

 

[StatusX]

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

 

[precondition]

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

 

[Dragonfall]

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'

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.

 

[istevenson]

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'

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.

 

[HallsofIvy]

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?

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.