<s>[INST] 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.
 [/INST] 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
$$\sin^2(2t)\le 1 \iff $$
$$4\sin^2(t)\cos^2(t)\le (\cos^2(t)+\sin^2(t))^2\iff $$
$$4r^4\sin^2(t)\cos^2(t)\le r^4(\cos^2(t)+\sin^2(t))^2\iff$$
$$4x^2y^2\le (x^2+y^2)^2$$
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