Proof of Calculus III: Proving Limit as (x,y) -> (0,0) = 0

[Mona1990]

Hi!
I was wondering if someone could give me a couple hints on how to tackle the following proof!

Let f(x,y)= [ (lxl ^a)(lyl^b) ]/ [(lxl^c) + lyl^d] where a,b,c,d are positive numbers.
prove that if (a/c) + (b/d) > 1
then limit as (x,y) -> (0,0) of f(x,y) exists and equals zero.

thanks!

 

[Office_Shredder]

The limit as (x,y) goes to (0,0) of f(x,y) is zero if and only if the limit (x,y) -> (0,0) of |f(x,y)| =0

Since |f(x,y)| is always positive, this is equivalent to saying

[tex] \lim_{(x,y) \rightarrow (0,0)} \frac{1}{|f(x,y)|} = \infty[/tex]

This works for general f(x,y). In this case, we know f(x,y) > 0 if (x,y) is not equal to (0,0) so you just need to show[tex] \lim_{(x,y) \rightarrow (0,0)} \frac{1}{f(x,y)} = \infty[/tex]

And this is easier since you can split up the numerator and start comparing a to c and b to d

 

[Mona1990]

Hey!
thanks a lot :) makes sense now!