$$\lim_{(x,y)\to(0,0)} \frac{\sqrt{|y|}\sin^3(x)}{(x^2+y^2)^{1.5}}$$
My thoughts and work:
First of all I multiplied and dividing by $x^3$, so I can get rid of the $\sin^3x$.
$$\lim_{(x,y)\to(0,0)}\frac{\sin^3(x)}{x^3} \frac{\sqrt{|y|}x^3}{(x^2+y^2)^{1.5}}$$
Now I know the left fraction goes to $1$.
And by intuition, I can see that the right fraction goes to $0$, since the power of the numerator is $3.5$. and of the denominator $3$.
But I'm struggling of how to prove it mathematically.
Any help is appreciated, thanks in advance!
Best Answer
$$\frac {|x^{3}|\sqrt {|y|}} {(x^{2}+y^{2})^{1.5}} \leq \frac {|x^{3}|\sqrt {|y|}} {(x^{2})^{1.5}}=\sqrt {|y|} \to 0.$$