Multivariable Calculus: When to use Squeeze Theorem to solve limits?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 7K views
theBEAST
Messages
361
Reaction score
0

Homework Statement


For example in:
lim (x,y) -> (0,0) [(xy^2)/(x^2+y^4)]

This limit does not exist (according to textbook), but if you use squeeze theorem since y^2<(x^2+y^4)
y^2/(x^2+y^4) <= 1 and therefore

0 <= (xy^2)/(x^2+y^4) <= x
as x--> 0
so
lim (x,y) -> (0,0) [(xy^2)/(x^2+y^4)] = 0?
 
Physics news on Phys.org
What happens if you look at y=sqrt(x)?
 
dirk_mec1 said:
What happens if you look at y=sqrt(x)?

I see what you mean but what if i was looking at a different limit and I got 0 every time while using different paths. I still couldn't prove it with squeeze theorem because what if there is just this one path that i never tried.