<s>[INST] Limit of multivariable function $f(x,y) = {(x^2+y^2)}^{x^2y^2}$ $$f(x,y) = {(x^2+y^2)}^{x^2y^2}$$
I need to find the limit at (0,0) point
I applied the exponent rule and got $$e^{x^2y^2ln(x^2+y^2)}$$
and now with chain rule, I need to find the limit of $${x^2y^2ln(x^2+y^2)}$$
and how? :D
There isn't L'Hôpital's rule for multivariable function, right?
 [/INST] For  $0 < x^2+y^2<1$ we have
$$1 \geqslant (x^2+y^2)^{x^2y^2} \geqslant (x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}$$
and
$$\lim\limits_{x \to 0 \\y \to 0}(x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}=\lim\limits_{t \to 0+}t^{\frac{1}{4}t^2} = 1$$
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