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2-variable function limit

  • #1
[itex]lim_{(x,y)\rightarrow(0,0)}\frac{y^{3}}{x^{4}+sin^{2}y}[/itex]

I need help with this above

any suggestions?

i dont know how to do this and have exam tomorrow :(
 

Answers and Replies

  • #2
please im short of time......

i would be very grateful
 
  • #3
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I'll give you my take on this

Note that

[tex]x^4 \geq 0 [/tex]

And that

[tex]0\leq sin^2 y \leq 1[/tex]

Then we know that

[tex]x^4 \leq sin^2 y + x^4 \leq 1 + x^4[/tex]

Also,all of this is just to tell you that you may have to use the squeeze theorem to solve this.
 
  • #4
LCKurtz
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Try a couple of different paths, like letting y → 0 first or x → 0 first to see if perchance they are different.
 
  • #5
[tex]x^4 \leq sin^2 y + x^4 \leq 1 + x^4[/tex]
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 doesnt take me any step further :(

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 [itex]lim_{x\rightarrow0}(lim_{y\rightarrow0}A)[/itex] , where A =

[itex]\frac{y^{3}}{x^{4}+sin^{2}y}[/itex]
 
  • #6
LCKurtz
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Do you mean calculating [itex]lim_{x\rightarrow0}(lim_{y\rightarrow0}A)[/itex] , where A =

[itex]\frac{y^{3}}{x^{4}+sin^ {2}y}[/itex]
Yes. And the reverse order too. What can you conclude if they come out not equal to each other?
 
  • #7
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 dont know how to calculate these limit (d'hospital) ??
 
  • #8
LCKurtz
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That the limit as (x,y)->(0,0) does not exist?
That's correct.
but i dont 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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