Limit of multivariable function $f(x,y) = {(x^2+y^2)}^{x^2y^2}$

Question

$$f(x,y) = {(x^2+y^2)}^{x^2y^2}$$

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?

Answer 1

Hint:

Set $x=r\cos(t)$ and $y=r\sin(t)$ and take $r\rightarrow 0$

Answer 2

Observe immediately that the function is symmetric: that is $$f(x,y)=f(y,x)$$

Then, you can set $y=y(x)$ as any function so that $x\to 0\implies y\to 0$ and calculate $$\lim_{x\to 0}f(x,y(x))$$ as a limit of one variable.

Some examples of $y(x)$ are $y(x)=0,y(x)=kx, k\in \Bbb R^+$

Answer 3

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 $$\lim\limits_{x \to 0 \\y \to 0}(x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}=\lim\limits_{t \to 0+}t^{\frac{1}{4}t^2} = 1$$

Answer 4

We have that

$$ {(x^2+y^2)}^{x^2y^2}=e^{x^2y^2 \log(x^2+y^2)} \to 1$$

indeed

$$x^2y^2 \log(x^2+y^2)=(x^2+y^2) \log(x^2+y^2) \frac{x^2y^2}{x^2+y^2} \to 0\cdot 0=0 $$

since by $t=x^2+y^2 \to 0$ by standard limits

$$(x^2+y^2) \log(x^2+y^2) =t\log t \to 0$$

and since $x^2+y^2 \ge 2xy$

$$0\le\frac{x^2y^2}{x^2+y^2} \le \frac{x^2y^2}{2xy} =\frac12 xy \to 0$$