lim(x,y)-(0,0) of ((5y^4)*cos^2(x))/(x^4+y^4) I tried to use the squeeze theorem, arguing that x^4+y^4 = y^4 so (1/(x^4+y^4)) 0)5*cos^2(x) = 5, which gives me that 0

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lim(x,y)->(0,0) of ((5y^4)*cos^2(x))/(x^4+y^4) I tried to use the squeeze theorem, arguing that x^4+y^4 >= y^4 so (1/(x^4+y^4)) <= 1/y^4, multiply through by abs((5y^4)*cos^2(x)) which yields lim(x->0)5*cos^2(x) = 5, which gives me that 0<=lim(original)<=5 This doesn't seem like sufficient evidence to say that the limit doesn't exist, but I'm not really sure where to take it.

perl · Answer

Letting x = 0 yields 
  lim(y->0) 5y^4 / (0 + y^4)
= lim (y->0) 5*1
= 5

but letting y = 0 yields
  lim ( x->0) ((5*0^4)*cos^2(x))/(x^4+0^4) 
= lim (x->0) 0/x^4
= lim (x->0) 0
= 0

perl · Answer

since the limits do not match , (it is not path independent) the limit does not exist

anonymous · Answer

Thanks! I can't believe I didn't notice that. Guess I got hung up on the Squeeze Theorem/Polars.

perl · Answer

:)

perl · Answer

and do you see why 
lim (x->0)  0/x^4  = 0 ?

perl · Answer

because in the limit, the actual value at x = 0 does not matter. 
so for x not equal to zero, 0/x^4 or 0/x  goes to zero

perl · Answer

so we wouldn't use Lhopitals, in fact that would make matters worse :)

perl · Answer

for example 0/.000001 is still zero , no matter how small you make the denominator

perl · Answer

@ganeshie8  if you have time can you look this over to check for any mistakes 

Ideally we could graph the surface and see why the limit does not exist.

jhannybean · Answer

well, you can try several paths, like $x=y$ and $y=x^2$ etc.

ganeshie8 · Answer

yes we can pick any two simple paths
 showing the limit is different along two paths is sufficient

ganeshie8 · Answer

x=0 and y=0 paths will do, it looks good to me :)

ganeshie8 · Answer

$$\lim \limits_{t\to 0 } ~\dfrac{0}{t} = \lim \limits_{t\to 0 }~0 = 0 $$

ganeshie8 · Answer

lim(x,0)->(0,0) of ((5y^4)*cos^2(x))/(x^4+y^4) = 0

lim(0,y)->(0,0) of ((5y^4)*cos^2(x))/(x^4+y^4) = 5

perl · Answer

right, i made a typo 
i should have had 
lim (x,0) -> (0,0) 

interestingly if x = y the limit is 5/2

perl · Answer

lim(x,x)->(0,0) ((5y^4)*cos^2(x))/(x^4+y^4) =5/2
lim(y,y)->(0,0) ((5y^4)*cos^2(x))/(x^4+y^4) =5/2

ganeshie8 · Answer

nice there could be other paths along which the limit might exist