Two variables calculus-not too hard

• Mar 14th 2010, 07:56 AM
WannaBe
Two variables calculus-not too hard
1. Given the function:
$\displaystyle f(x,y) = \frac{xy^{2}}{x^{4}+y^{2}}$
Does the limit $\displaystyle \lim_{(x,y)\to(0,0)} f(x)$ exist?

2. For positive $\displaystyle \alpha , \beta$, show that:
$\displaystyle \lim_{(x,y)\to(0,0)} x^{\alpha}y^{\beta}ln(x^{4}+y^{2}) =0$

I'll be delighted to get some guidance in these two...
In the first one I've tried to devide the function by $\displaystyle y^{2}$, but I couldn't understand how to continue...
About the second one- I think I should do something wil l'hospital's law or somehing... But I can't figure out how...

Thanks!
• Mar 14th 2010, 08:15 AM
Miss
Quote:

Originally Posted by WannaBe
1. Given the function:
$\displaystyle f(x,y) = \frac{xy^{2}}{x^{4}+y^{2}}$
Does the limit $\displaystyle \lim_{(x,y)\to(0,0)} f(x)$ exist?

2. For positive $\displaystyle \alpha , \beta$, show that:
$\displaystyle \lim_{(x,y)\to(0,0)} x^{\alpha}y^{\beta}ln(x^{4}+y^{2}) =0$

I'll be delighted to get some guidance in these two...
In the first one I've tried to devide the function by $\displaystyle y^{2}$, but I couldn't understand how to continue...
About the second one- I think I should do something wil l'hospital's law or somehing... But I can't figure out how...

Thanks!

For the first, use the sandwich theorem:

$\displaystyle \left| \frac{xy^2}{x^4+y^2} \right| \leq \frac{xy^2}{y^2}=x$.