How can the limit of this function be shown to be 2 as x and y approach 0?

[oahsen]

Homework Statement

as x,y goes to 0,0 lim(2x/(x^2+x+y^2)

The Attempt at a Solution

It is obvious that the limit is 2 but I can not show it. I tried to change polar coordinates etc. but nothing did not help. What should I do?

 

[Mathgician]

I would suggest doing limits for each varable separately

make either x=0 or y=0 and do the limits separately, or in other words, the limit in the y direction or x direction.

 

[nightowl03d]

First take the limit as y goes to 0, to get to
lim(2x/(x^2+x)).
Divide out the common factor of x to get
lim(2/(1+x))
Taking this limit will give 2.

 

[matt grime]

You can't do that. the point is that it has to be independent of the path chosen. You cannot let y tend to zero, then x tend to zero - this is just taking a path along the x axis.

The best advice I heard was to always convert to polars and then let r tend to zero.

Let t be theta then we have

2rcos(t)/(r^2+rcos(t)) = 2cos(t)/(r+cos(t))

As r tends to zero that tends to 2, as long as cos(t) is not zero.

this of course shows that the limit does not exist, contrary to what the OP asked. But then that was obvious anyway: on the y-axis (so x=0) away from y=0 that function is identically zero.

 

[Mathgician]

First take the limit as y goes to 0, to get to
lim(2x/(x^2+x)).
Divide out the common factor of x to get
lim(2/(1+x))
Taking this limit will give 2.

Yes, this is correct, and in the Y direction the limit is negligible because it goes to zero and the limit goes to zero if you ignore the x direction.

 

[HallsofIvy]

First take the limit as y goes to 0, to get to
lim(2x/(x^2+x)).
Divide out the common factor of x to get
lim(2/(1+x))
Taking this limit will give 2.

Yes, this is correct, and in the Y direction the limit is negligible because it goes to zero and the limit goes to zero if you ignore the x direction.

OK, if "y goes to 0" first, you get 2x/(x²+ x)= 2/(x+1) which goes to 2 as x goes to 0. If x goes to 0 first, you get 0/y = 0 which goes to 0 as y goes to 0. What does that tell you about the limit itself?

 

[Mathgician]

corrections: limit does not exist for f(x,y), because limit in the x direction = 2 and in the y = 0.

 

[nightowl03d]

You can't do that. the point is that it has to be independent of the path chosen. You cannot let y tend to zero, then x tend to zero - this is just taking a path along the x axis.

The best advice I heard was to always convert to polars and then let r tend to zero.

Let t be theta then we have

2rcos(t)/(r^2+rcos(t)) = 2cos(t)/(r+cos(t))

As r tends to zero that tends to 2, as long as cos(t) is not zero.

this of course shows that the limit does not exist, contrary to what the OP asked. But then that was obvious anyway: on the y-axis (so x=0) away from y=0 that function is identically zero.

Quite right, the limit should be the same along all dimensions under consideration or it doesn't exist.