Limit of $(1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}}$, where do I get it wrong? I am tasked with evaluating
$$\lim_{(x, y) \to (0, 0)} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}}.$$
We can rewrite the expression it as follows:
$$
e^{\ln(1+x^2+y^2)\cdot\frac{1}{x^2+y^2+x y^2}}
$$
Now, my first guess was that this tends to $1$.  But that is apparently incorrect, it tends towards $e$, which one can see just by graphing it. That is also what the book says.
Now, we have the "standard limit":
$$\lim_{x \to 0^+} \frac{\ln(1+x)}x = 1$$
So I can kinda see how we can get the power to tend toward $1$ so we get $e^1$, but the expression is not exactly the same, we have $xy^2$ tacked on to the end of the denominator of the fraction. Furthermore, $xy^2$ doesn't have to be positive.
I tried using the substitution $x=\frac{1}{y^2}$, that gives us an expression that tends towards infinity both at $0$ and infinity and is never zero.
Can anyone point out to me what I'm missing. It feels like it should be super-obvious. 

Let $x^2+y^2=r^2 .$ We have $$|x|<a\implies |x y^2|\leq a y^2\leq a(x^2+y^2)=a r^2 .$$ So let $xy^2=r^2(1+b)$ where $b\to 0$ as $r\to 0 .$ Observe that this implies that $x^2+y^2+xy^2>0$ for sufficiently small positive $r.$ So, for all sufficiently small $r>0$ we have $$\log (1+x^2+y^2)^{1/(x^2+y^2+x y^2)}=$$  $$ =\frac {1} {1+b} \frac {1}{r^2}\log (1+r^2)$$ which tends to $1$ as $r\to 0.$  
