- #1

- 4

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I need help with this above

any suggestions?

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

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- Thread starter maciejewski
- Start date

- #1

- 4

- 0

I need help with this above

any suggestions?

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

- #2

- 4

- 0

please im short of time......

i would be very grateful

i would be very grateful

- #3

- 2,571

- 1

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

- 9,557

- 767

- #5

- 4

- 0

[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 :(

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

- 9,557

- 767

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

- 4

- 0

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

- 9,557

- 767

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

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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