$\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
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.