How do you find the following multivariable limit? How do you find the following limit:
$$\lim_{x,y\to 0} \frac{|x||y|^4}{|x|^4+|y|^5}$$
I've tried to apply the squeeze theorem, but unlike a lot of other questions I encounter, the degree of $x$ in the numerator is too small for me to write the function as a constant times a function that goes to zero as $x,y \rightarrow 0$.
 A: We can rewrite the expression like so
$$\frac{|x|y^4}{x^4+|y|^5} = \frac{|x|y^4}{\sqrt{x^4+|y|^5}}\cdot\frac{1}{\sqrt{x^4+|y|^5}}$$
By AM-GM inequality (or if you don't want to use AM-GM, simply complete the square) we have that
$$x^4+|y|^5 \geq 2x^2|y|^{\frac{5}{2}}$$
which means
$$\frac{|x|y^4}{\sqrt{x^4+|y|^5}}\cdot\frac{1}{\sqrt{x^4+|y|^5}} \leq \frac{|x|y^4}{\sqrt{2x^2|y|^{\frac{5}{2}}}}\cdot\frac{1}{\sqrt{x^4+|y|^5}} = \frac{|y|^{\frac{11}{4}}}{\sqrt{2}\sqrt{x^4+|y|^5}}$$
and lastly we have that
$$\frac{|y|^{\frac{11}{4}}}{\sqrt{2}\sqrt{x^4+|y|^5}} \leq \frac{|y|^{\frac{11}{4}}}{\sqrt{2}|y|^{\frac{5}{2}}} = \left|\frac{y}{4}\right|^{\frac{1}{4}}$$
thus we have that the limit
$$\lim_{(x,y)\to(0,0)}\frac{|x|y^4}{x^4+|y|^5} \leq \lim_{(x,y)\to(0,0)} \left|\frac{y}{4}\right|^{\frac{1}{4}} = 0$$
by squeeze theorem. As a rule of thumb, if the path substitution that makes the denominator homogeneous (i.e. the sum of terms of the same degree) has the limit exist and agree with the other paths you've tried, then it probably means the limit exists, and more importantly, a dominating function exists.
A: We have that by $|y|=|v|^{\frac45}\to 0$
$$\frac{|x||y|^4}{|x|^4+|y|^5} =\frac{|x||v|^{\frac{16}5}}{|x|^4+|v|^4} \to 0$$
as we can conclude by polar coordinates, indeed
$$\frac{|x||v|^{\frac{16}5}}{|x|^4+|v|^4}=r^{\frac15}\frac{|\cos \theta||\sin\theta|^{\frac{16}5}}{|\cos \theta|^4+|\sin \theta|^4} \to 0$$
since $\forall \theta$
$$|\cos \theta|^4+|\sin \theta|^4 \ge c >0$$
