How to Calculate a 2-Variable Function Limit?

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lim_{(x,y)\rightarrow(0,0)}\frac{y^{3}}{x^{4}+sin^{2}y}

I need help with this above

any suggestions?

i don't know how to do this and have exam tomorrow :(
 
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please I am short of time...

i would be very grateful
 
I'll give you my take on this

Note that

x^4 \geq 0

And that

0\leq sin^2 y \leq 1

Then we know that

x^4 \leq sin^2 y + x^4 \leq 1 + x^4

Also,all of this is just to tell you that you may have to use the squeeze theorem to solve this.
 
Try a couple of different paths, like letting y → 0 first or x → 0 first to see if perchance they are different.
 
flyingpig said:
x^4 \leq sin^2 y + x^4 \leq 1 + x^4
Also,all of this is just to tell you that you may have to use the squeeze theorem to solve this.

thanks for a tip, but I've already figured this and it probably doesn't take me any step further :(

LCKurtz said:
Try a couple of different paths, like letting y → 0 first or x → 0 first to see if perchance they are different.

Do you mean calculating lim_{x\rightarrow0}(lim_{y\rightarrow0}A) , where A =

\frac{y^{3}}{x^{4}+sin^{2}y}
 
maciejewski said:
Do you mean calculating lim_{x\rightarrow0}(lim_{y\rightarrow0}A) , where A =

\frac{y^{3}}{x^{4}+sin^ {2}y}

Yes. And the reverse order too. What can you conclude if they come out not equal to each other?
 
LCKurtz said:
Yes. And the reverse order too. What can you conclude if they come out not equal to each other?

That the limit as (x,y)->(0,0) does not exist?

but i don't know how to calculate these limit (d'hospital) ??
 
maciejewski said:
That the limit as (x,y)->(0,0) does not exist?
That's correct.
but i don't know how to calculate these limit (d'hospital) ??

Have you tried anything? What happens if y → 0 first? What happens if x → 0 first?
 
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