Solving limit - $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$ $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
Since $x$ approaches $0$ and $y$ also approaches $0$ we can suspect that $0<x^2 + y^2<1$. For every $x,y\in\Bbb R$, we have that $\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$.
Now, $1\geq (x^2+y^2)^{x^2y^2}\geq (x^2+y^2)^{\frac{1}{4}(x^2 + y^2)^2}$, then substitute $(x^2 + y^2)=t$
$1\geq \lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}\geq \lim_{t\to0}t^{\frac{1}{4}t^2}=\lim_{t\to0}e^{\frac{1}{4}t^2\ln t}=e^0=1$
This is how my professor solved this limit. What I don't understand is this part:
$\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$
How can I prove it? And this would never come to my mind, is there maybe some other way to solve the limit? Grateful in advance.
 A: hint
The difference gives
$$\frac 14(x^2+y^2)^2-x^2y^2=$$
$$\frac 14\Bigl(x^4+y^4+2x^2y^2-4x^2y^2\Bigr)=$$
$$\frac 14\Bigl(x^4+y^4-2x^2y^2\Bigr)=$$
$$\frac 14(x^2-y^2)^2\ge 0$$
Other proof
Putting
$$x=r\cos(t)\;,\;y=r\sin(t)$$
We know that
$$\sin^2(2t)\le 1 \iff $$
$$4\sin^2(t)\cos^2(t)\le (\cos^2(t)+\sin^2(t))^2\iff $$
$$4r^4\sin^2(t)\cos^2(t)\le r^4(\cos^2(t)+\sin^2(t))^2\iff$$
$$4x^2y^2\le (x^2+y^2)^2$$
A: Here's another approach to this limit. Notice for any $(x,y)\neq (0,0)$ that $$
x^2y^2 \ln(x^2+y^2)=f(x,y)(x^2+y^2)\ln(x^2+y^2)$$ where $f(x,y)=\frac{x^2y^2}{x^2+y^2}$. Clearly $(x^2+y^2)\ln(x^2+y^2)\rightarrow 0$ as $(x,y)\rightarrow (0,0)$ while $f(x,y)$ is bounded on the punctured disc $x^2+y^2<1$, $(x,y)\neq (0,0)$. To see this, observe $f(x,0)=0$ for $x\in(-1,0)\cup(0,1)$ and for $x^2+y^2<1,y\neq 0$ we have $$\bigg|\frac{x^2y^2}{x^2+y^2}\Bigg|\leq \Bigg|\frac{x^2y^2}{0+y^2}\Bigg|=x^2\leq x^2+y^2<1$$ This shows $f$ is bounded above by $1$ on the punctured disc, making $$\lim_{(x,y)\rightarrow (0,0)} x^2y^2\ln(x^2+y^2)=0$$ Finally, $$\lim_{(x,y)\rightarrow (0,0)}(x^2+y^2)^{x^2y^2}=\lim_{(x,y)\rightarrow (0,0)}e^{x^2y^2\ln(x^2+y^2)}=e^0=1$$
