Another Day[9] Question

x/y as they both approach infinity is undefined
how do we prove =/= 2x or y=/= 0.001x

same applies to x1 and x2 having the same support does not mean they are equal

edit * fuck me meant to say undefined

In all of these you have to be careful about how you specify the limit.

Take x/y for example.

One way to specify them going to infinity is that the distance from (x,y) to the origin goes to infinity (ie x^2 + y^2 is going to infinity). This will lead to a divergent limit. Look at the below graph:

https://www.wolframalpha.com/input/?i=f(x,y)+%3D+x%2Fy

You see how near the x-axis f(x,y) = x/y is really big, but near the y-axis it is close to 0.


You could make x go to infinity first then y and get the limit to be infinity. You could make y go to infinity first then make x go to infinity to get 0. (think what you get when you do the first limit).

Or maybe y goes to infinity like 2x. Then the limit is 1/2.



For the second question similarly you need to specify how it goes to infinity.

On April 14 2017 23:33 failsafe wrote: it seems the limit of x/x would be 1 as x goes to infinity. but what if we wrote x/y and the limit of both of those to infinity. then 1? but now x1 in n and x2 in r finally x1/x2 are we able to do this? cannot only in r?

On April 14 2017 23:33 failsafe wrote: it seems the limit of x/x would be 1 as x goes to infinity. but what if we wrote x/y and the limit of both of those to infinity. then 1? but now x1 in n and x2 in r finally x1/x2 are we able to do this? cannot only in r?

Whether one function is in R and the other in N is irrelevant for this.

When you have a fraction and the limits of both numerator and denominator are infinity, then you can take the fraction of their derivatives (L'Hopital's rule).

The idea is to compare the rate of growth of the functions. And derivatives tell us the rate of growth of a function.

So for example:
x1 = x (x in N)
x2 = e^x (e^x in R)
then lim x1/x2 = lim x/e^x = lim deriv(x)/deriv(e^x) = lim 1/e^x = 0.

Both functions go to infinity, but denom. goes there strictly faster, thus limit is 0.

On April 15 2017 06:35 KoeBawlt wrote: In all of these you have to be careful about how you specify the limit. Take x/y for example. One way to specify them going to infinity is that the distance from (x,y) to the origin goes to infinity (ie x^2 + y^2 is going to infinity). This will lead to a divergent limit. Look at the below graph: https://www.wolframalpha.com/input/?i=f(x,y)+%3D+x%2Fy You see how near the x-axis f(x,y) = x/y is really big, but near the y-axis it is close to 0. You could make x go to infinity first then y and get the limit to be infinity. You could make y go to infinity first then make x go to infinity to get 0. (think what you get when you do the first limit). Or maybe y goes to infinity like 2x. Then the limit is 1/2. For the second question similarly you need to specify how it goes to infinity.

yeah, shit dude. wolfram is fucking hacks. my mind is blown.

i met this guy several years ago while i was in a psych ward who explained to me that 1*1 and 1/1 defied math because the functional operators couldn't obtain the result that:

1*1 = 1

and

1/1 = 1

i feel like that must be true after looking at the wolfram graph.

obviously 1 =/= 1

i feel like humans are meant to be eaten alive by trees and that my house is going to murder me

this seems more and more true. i diligently sprayed my computer desk area with lysol for several days and ultimately ended up paying almost $800 in apartment repair fees. this was after purchasing a herman miller aeron.

1 most definitely equals 1

Joe answered your question very well

On April 15 2017 17:47 traxamillion wrote: 1 most definitely equals 1

lol, yeah but if we come to the same conclusion for the same reason we'll both lose.