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.
 A: 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 $(x,y)\to(0,0)$.
A: Cauchy inequality (arithmetic means larger or equal to geometric mean) implies that for every $a,b\ge0$
$$
\underbrace{\frac{a}{k}+\cdots+\frac{a}{k}}_{k\,\,\text{times}}
+\underbrace{\frac{b}{\ell}+\cdots+\frac{b}{\ell}}_{\ell\,\,\,\text{times}}
\ge (k+\ell)\,\,\sqrt[k+\ell]{\left(\frac{a}{k}\right)^k\left(\frac{b}{\ell}\right)^\ell}
$$
or
$$
a+b\ge c_{k,\ell}a^{\frac{k}{k+\ell}}b^{\frac{\ell}{k+\ell}}
$$
for a suitable constant $c_{k,\ell}>0$.
So in our case ($a=x^4$ and $b=y^8$)
$$
x^4+y^8\ge c_{3,5}(x^4)^{3/8}(y^8)^{5/8}=c_{3,5}\lvert x\rvert^{3/2}\lvert y\rvert^5,
$$
or
$$
(x^4+y^8)^2\ge c_{3,5}^2\lvert x\rvert^3 \lvert y\rvert^{10}
$$
and hence
$$
\lvert f(x,y)\rvert=\frac{4\lvert x^3y^{11}\rvert}{(x^4+y^8)^2}\le \frac{\lvert y\rvert}{c_{3,5}^2}.
$$
Clearly
$$
\lim_{(x,y)\to(0,0)}f(x,y)=0.
$$
A: 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.
A: 
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
$$
\frac{3}4+\frac{11}8=\frac{17}8>2.
$$
A: Show that the limit is 0 as $(x,y)\to(0,0)$ in the first quarter $\{x>0,y>0\}$. Assuming the opposite we get such $x_n,y_n>0 (n\in\mathbb{N})$ tending to 0 and $A>0$ that
$$\frac{x_n^{1.5}y_n^{5.5}}{x_n^{4}+y_n^8}>\frac{1}{A}$$
or
$$\frac{x_n^{2.5}}{y_n^{5.5}}+\frac{y_n^{2.5}}{x_n^{1.5}}<A.$$
Whence
$$\frac{x_n^{2.5}}{y_n^{5.5}}<A,\quad \frac{y_n^{2.5}}{x_n^{1.5}}<A,$$
which yield (for som $A'$ and $A''$)
$$x_n<A' y_n^{11/5}<A'' x_n^{33/25},$$
then
$$1<A'' x_n^{8/25}.$$
Since $x_n\to 0$ we get a contradiction.
A: This isn't quite a complete answer, but it may help. Suppose we approach the origin along the path $y=x^a$. Then we obtain
$$\lim_{x\to 0} \frac{4x^{11a+3}}{(x^4+x^{8a})^2} = \lim_{x\to 0} \frac{4x^{11a+3}}{x^8+2x^{8a+4}+x^{16a}}$$
The only way this won't go to $0$ is if each exponent on bottom is greater than the exponent on top. That means you need:
$\begin{align}11a+3&<8 \\
11a+3&<8a+4 \\
11a+3&<16a
\end{align}$
These inequalities are inconsistent (in particular, the second and third contradict each other), so you'll get a limit of $0$ approaching along any path of the given form. I suspect the limit is $0$, but I'm not yet sure how to prove it.
