Why does 1/x approach 0 as x approaches infinity?

[Josh S Thompson]

No. ϵ\epsilon is a particular (positive, and usually close to 0) number that someone tells you. Your part of the dialog is to find a positive number δ\delta so that some condition is met.

If the other person is satisfied that you have established the limit, you're done. If not, the other person will make it harder, by telling you a smaller positive number, ϵ\epsilon. You then have to find a corresponding, but possibly different value of δ\delta. The process continues until the other person finally gives up.

Wow, thank you that is a great explanation,
There are multiple turns in the game and it repeats at the beginning of the statement.

But what if there exists goes first, would you just have to prove the statement for one delta and all epsilon and the game would not repeat.

 

[Mark44]

Wow, thank you that is a great explanation,
There are multiple turns in the game and it repeats at the beginning of the statement.

But what if there exists goes first, would you just have to prove the statement for one delta and all epsilon and the game would not repeat.

That's not how it works. You can't arbitrarily change the definition of $\lim_{x \to a} f(x) = L$, which goes like this:
"For each $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < |x - a| < \delta, |f(x) - L| < \epsilon$."
Sometimes you see this with more symbolism, but this captures the idea of the definition.

You start with $\epsilon$ that someone gives you, and you find $\delta$ so that when x is within $\delta$ of a, then f(x) is within $\epsilon$ of L.

 

[Helolo]

$1/x$ doesn't eaqual to $\infty$ if $x=0$ and $1/x$ doesn't equal to $0$ if $x=\infty$, but the limit of 1/x when x approach ∞ or x approach to 0 it does and by the way Josh S. Thompson u need to learn about Limitations again. ☺

 

[Josh S Thompson]

If the other person is satisfied that you have established the limit, you're done. If not, the other person will make it harder, by telling you a smaller positive number, ϵ\epsilon. You then have to find a corresponding, but possibly different value of δ\delta. The process continues until the other person finally gives up.

What if the person picking epsilon never gives up?

 

[micromass]

What if the person picking epsilon never gives up?

Then the game goes on forever. The point is that the person choosing the $\delta$ can never lose.

 

[Josh S Thompson]

Then the game goes on forever. The point is that the person choosing the $\delta$ can never lose.

thats why its a limit becasue the game goes on forever.