Why do polar coordinates give different answers for this limit problem?

[tohauz]

hi.
Spse you want to find limf(x,y) as (x,y)->0.
You can use polar coordinates and get
limf(rcost,rsint) as r->0. And these limits are same.
Now, i initially thought that
1)if limf(rcost,rsint)is independent of t,then limit
exists and is equal tolimf(x,y) as (x,y)->0.
2)If it depends on t, then limit DNE.
But, i did this limits using these methods and got different answers:
limit(xy^{4}/x^{2}+y^{8}) as (x,y)->0.
So if you try x=y^{4} path, you get 0.5
y=0 path, you get 0. So DNE
if you convert into polar, you get zero.

Where is the mistake??
Thanks in advance

 

[sin123]

I cannot tell you whether your limit exists or not, but I can tell you why you get different answers. In polar coordinates, when you take the limit as r -> 0 and t is fixed, you are "only" checking the limit along straight lines. It is equivalent to checking x = m y only.

I don't know whether or not that's sufficient for the limit to exist. But when you graph your function in close neighborhoods of (0,0), the limit seems to be (0,0) indeed.