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. 
 A: Your rewrite is a step in the right direction. Next, use polar coordinates. The limit becomes:
$$\lim_{\rho \to 0} \exp\left(\frac{\ln(1 + \rho^2)}{\rho^2(1 + \rho\sin^2\theta\cos\theta)}\right)$$
Using the fact that $\ln(1 + t) \sim_0 t$, and that $\sin^2\theta\cos\theta$ is bounded, you can easily finish.
A: The given limit exists if and only if the limit
$$
\lim_{(x,y)\to(0,0)}\frac{\ln(1+x^2+y^2)}{x^2+y^2+xy^2}
$$
exists. Pass to polar coordinates: $x=r\cos\varphi$, $y=r\sin\varphi$, so the limit becomes
$$
\lim_{r\to0}\frac{\ln(1+r^2)}{r^2(1+r\cos\varphi\sin^2\varphi)}=
\lim_{r\to0}\frac{\ln(1+r^2)}{r^2}
\lim_{r\to0}\frac{1}{1+r\cos\varphi\sin^2\varphi}=1
$$
A: 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 $r\to 0.$  
A: If we write $u = x^2+y^2,$ the expression equals
$$[(1+u)^{1/u}]^{u/(u+xy^2)} = [(1+u)^{1/u}]^{1/(1+(xy^2/u))} .$$
As $u\to 0,$ the expression inside the brackets $\to e.$ It's a simple argument to show $xy^2/u \to 0.$ Thus the exponent outside the brackets $\to 1,$ and the desired limit is $e^1=e.$
