URL: http://www.mathsisfun.com/forum/viewtopic.php?pid=61986

I can see your point of view: the "n"th term is very small and so can be dismissed. But could you not eliminate it all together (somehow)?

And back to 1/infinity, I may have trouble actually evaluating the function, but I can evaluate it's limit.

]]>However, you can check a similiar problem posted by Heldensheld in the Helpme section. Very Tricky!

]]>I can interpret he intended to evaluate the function of 1/x on infinity.

It may have been my intention, but ... did I succeed?

]]>There is no such thing as "on infinity". Functions are defined on the real numbers. Infinity is not a real number, and thus, not in the domain of functions.

Sorry Sir! I don't know you recently begin to censor the phrase " on infinity".

But I suppose as MIF has put infinity into the function 1/x, I can interpret he intended to evaluate the function of 1/x on infinity.

]]>Limits Page wrote:

]]>The method basically says

Where you have the example of (x²+1)/(x-1), it might be worth noting that as well as trying values that get progressively closer to 1 and looking for a pattern, you can also find the limit algebraically.

(x²+1)/(x-1) = (x+1)(x-1)/(x-1) = x+1 for all x ≠ 1. Therefore, as x approaches 1, then the function will approach 2.

If you want to get really advanced, you could even perhaps throw in a mention of L'Hopital's rule.

Also, at one point you have (1+1/n)^2, where the 2 should be an n.

It looks like it's going to be a very useful and interesting page when it's finished though, so keep up the good work!

(The method is certainly talkative. )

You can get a rare insight into my scattered thought processes as I put a page together

I am currently pushing my brain to show the formal definition in an easy way, through words, diagrams or animation. Inspiration will arrive I hope.

]]>Without further ado, here is Shadrach's Extension of Binary Operators to Infinity (SEBO for short):

So with this, we may remark that 1/infinity = 0 and 1^infinity = 1, but we can't say anything on what infinity / infinity is.

]]>It's OK, I chose the words "commonly accepted" carefully, and think a discussion would be enlightening ...

In fact my recent work on compound interest highlights an interesting thing.

A formula for "e" is

A naive interpretation of this would be

(1+1/∞)[sup]∞[/sup] => (1+0)[sup]∞[/sup] => 1

(

assuming (!)1/∞ = 0, and also that 1[sup]∞[/sup] = 1 which is interesting in its own right)But plugging in large values of n gives us e=2.7182... getting consistently more accurate.

Anyway, I will be writing this up somehow.

In the meantime feel free to tear this apart.

]]>

Yes, exactly. It carefully avoids mentioning Real Infinity. Or what happens on infinity.

There is no such thing as "on infinity". Functions are defined on the real numbers. Infinity is not a real number, and thus, not in the domain of functions.

]]>In fact my recent work on compound interest highlights an interesting thing.

A formula for "e" is

A naive interpretation of this would be

(1+1/∞)[sup]∞[/sup] => (1+0)[sup]∞[/sup] => 1

(*assuming (!)* 1/∞ = 0, and also that 1[sup]∞[/sup] = 1 which is interesting in its own right)

But plugging in large values of n gives us e=2.7182... getting consistently more accurate.

Anyway, I will be writing this up somehow.

In the meantime feel free to tear this apart.

]]>