Continuity of $f(x,y)=4x^3y^{11}(x^4+y^8)^{-2}$ at $(0,0)$

Well, the function is $$\frac{4x^3y^{11}}{(x^4+y^{8})^2}$$ and we want to know if it's continuous at $(0,0)$. I've tried as many trajectories as I could think of, and they all give $0$ as the limit. So I tried proving, by definition that its limit is in fact $0$, but to no avail.

Solution 1:

The limit at $(0,0)$ exists and is $0$.

To prove this, note that, for every $\theta$ in $(0,1)$ and every nonnegative $(u,v)$, $$u^\theta\cdot v^{1-\theta}\leqslant\max\{u,v\}^\theta\cdot\max\{u,v\}^{1-\theta}=\max\{u,v\}\leqslant u+v.$$ Thus, for every $(x,y)$, $$x^4+y^8\geqslant |x|^{4\theta}\cdot|y|^{8(1-\theta)},$$ which implies that $$ \left|\frac{x^3y^{11}}{(x^4+y^8)^2}\right|\leqslant |x|^{3-8\theta}\cdot|y|^{16\theta-5}. $$ The RHS goes to $0$ when $(x,y)$ goes to $(0,0)$ as soon as both exponents $3-8\theta$ and $16\theta-5$ are nonnegative and at least one of them is positive, that is, for every $\theta$ such that $\theta\geqslant\frac5{16}$ and $\theta\leqslant\frac38$. Since $\frac5{16}\leqslant\frac38$, this interval is not empty, which proves the result.

For example, $\theta=\frac13$ yields $$ \left|\frac{x^3y^{11}}{(x^4+y^8)^2}\right|\leqslant |x|^{1/3}\cdot|y|^{1/3}\to0. $$ More generally, for every positive $(a,b,c,d,e)$, $$ \frac{x^ay^b}{(x^c+y^d)^e}, $$ goes to $0$ at $(0,0)$ as soon as $$ \frac{a}c+\frac{b}d> e. $$ Note that $$ \frac{3}4+\frac{11}8=\frac{17}8>2. $$

Solution 2:

For all $(x,y)$ we have $$(x^4+y^8)^2\ge y^{16}\quad\hbox{and}\quad (x^4+y^8)^2\ge2x^4y^8\ .$$ Now suppose that $(x,y)\ne(0,0)\,$. If $|x|\le y^2$ we have $y\ne0$ and $$|f(x,y)|\le\frac{4|x|^3|y|^{11}}{|y|^{16}}\le4|y|\ ,$$ while if $|x|\ge y^2$ we have either $y=0$, when $f(x,y)=0$, or $y\ne0$ and $x\ne0$ and $$|f(x,y)|\le\frac{4|x|^3|y|^{11}}{2|x|^4|y|^8}=\frac{2|y|^3}{|x|}\le2|y|\ .$$ Therefore for all $(x,y)\ne(0,0)$ we have $$|f(x,y)|\le4|y|\ ,$$ and the right hand side tends to zero as $(x,y)\to(0,0)$.

Solution 3:

I'm adding this answer only because it brings out what is (to me) an interesting feature. First one can drop the $4$ and assume $x,y$ are positive, if one is claiming the limit is zero. Define $p(x,y)=x^3y^{11}/(x^4+y^8)^2$ and note we can split off one factor of $y$ and write $p(x,y)=y\cdot u(x,y)^2,$ where $$u(x,y)=\frac{\sqrt{x}^3y^5}{x^4+y^8}.$$ If one can show $u$ is bounded above we're done. But $u$ may, after dividing numerator and denominator by the numerator of it, be written in terms only of the ratio $t=\sqrt{x}/y$ as $u=1/(t^5+t^{-3}).$ This has by one variable methods its maximum at $t=(3/5)^{1/8}$ which makes $u$ about $0.51604,$ so $u$ is bounded as desired.