Wolframalpha it says the limit does not exist

[annoymage]

Homework Statement

use definition for the limit of a function to show that
$$\lim_{(x,y)\rightarrow (0,0)} xcos\frac{1}{y}$$

Homework Equations

n/a

The Attempt at a Solution

fisrt, i assumed it the limit is 0(i don't know if it is true), but i showed it, but something bugging me, i put in wolframalpha it says the limit does not exist, http://www.wolframalpha.com/input/?i=lim+xcos(1/y),(x,y)->(0,0)

so who is wrong here?

 

[lanedance]

at a quicklook
|cos(1/y)|<=1 for all y's,

so you can choose x small enough to bring you close enough to zero. The function will oscillate with increasing frequency, but with decreasing magnitude as you approach the origin

 

[annoymage]

at a quicklook
|cos(1/y)|<=1 for all y's,

so you can choose x small enough to bring you close enough to zero. The function will oscillate with increasing frequency, but with decreasing magnitude as you approach the origin

so, it really approaches to 0, but how come wolfram-alpha say it does not exist. did i wrote anything wrong?

 

[gomunkul51]

A limit of a function of several variables fails to exist if there are different limits from "different direction" i.e. from the "x" and "y" directions:

Thus, you need to check if the limit as x -> 0 (treat y as a constant) of the function is the same as the limit as y -> 0 (treat x as a constant).

Let's check limit from x direction:

lim[x->0] x*cos(1/y) = 0*cos(const.) = 0

But there is a problem with the limit of cos(1/y) when y->0.

http://www.wolframalpha.com/input/?i=lim+cos(1/y),+y->0

do you know why there is a problem with that y->0 limit? :)

Good Luck.

 

[lanedance]

yeah its an interesting one...

cos(1/y) is cleary undefined anywhere on the line y=0, so no limit on that line will exist for any function of cos(1/y)

however, it does have the property that as you get close to x=0, |x||cos(1/y)|<=|x|which tends to zero

also just to add, checking 2 startight line directions is suffcient to show a limit dne, but not to prove it exists