Limits: positive c so f(x,y)= (4x3-8x2y) / (cxy+9x2+9y2) converges as...

[fetante]

I have another problem:

I want to decide for which constants c, the function f(x,y)= (4x³-8x²y) / (cxy+9x2+9y²)

converges when (x,y)->(0,0)? And c must be positive.


How should I tackle this problem?

I try to use polar coordinates: x = rcos(Θ), y = rsin(Θ),

f(r) = (4r³cos³(Θ) - 8r³cos²(Θ)sin(Θ)) / (cr²cos(Θ)sin(Θ) + 9r^{2)
Simplifying}
f(r) = r³(4cos³(Θ) - 8cos²(Θ)sin(Θ)) / r²⁽ccos(Θ)sin(Θ) + 9)
^Simplifying
f(r) = r(4cos³(Θ) - 8cos²(Θ)sin(Θ)) / (ccos(Θ)sin(Θ) + 9)

^{And here I'm stuck. Is this even the right way to do it?}

 

[Ishuda]

I have another problem:

I want to decide for which constants c, the function f(x,y)= (4x³-8x²y) / (cxy+9x2+9y²)

converges when (x,y)->(0,0)? And c must be positive.


How should I tackle this problem?

I try to use polar coordinates: x = rcos(Θ), y = rsin(Θ),

f(r) = (4r³cos³(Θ) - 8r³cos²(Θ)sin(Θ)) / (cr²cos(Θ)sin(Θ) + 9r^{2)
Simplifying}
f(r) = r³(4cos³(Θ) - 8cos²(Θ)sin(Θ)) / r²⁽ccos(Θ)sin(Θ) + 9)
^Simplifying
f(r) = r(4cos³(Θ) - 8cos²(Θ)sin(Θ)) / (ccos(Θ)sin(Θ) + 9)

^{And here I'm stuck. Is this even the right way to do it?}

Continuing we then have
\(\displaystyle f(r,\theta)\,=\, \dfrac{8 r cos^2(\theta)[cos(\theta)-sin(\theta)]}{c\, sin(2\,\theta) +\, 18}\)
where I have used the double angle for the sine. Since c is positive
\(\displaystyle f(r,\theta)\,=\, \dfrac{8 r cos^2(\theta)[cos(\theta)-sin(\theta)]}{c\, [sin(2\,\theta) +\, \frac{18}{c}]}\)
which is a perfectly well behaved function except along the ray
\(\displaystyle \theta\,=\, \dfrac{1}{2}sin^{-1}\,\left(\dfrac{18}{c}\right)\)
Suppose c were greater than or equal to 18? Suppose it were less than 18 (and greater than zero)?