Proving that $\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}=1$ using polar coordinates Am I doing this right? I rewrite the function as follows:
$$(r^2\cos^2\theta+r^2\sin^2\theta)^{r^4\cos^2\theta\sin^2\theta} \stackrel{\text{various trig identities}}{=} r^{\frac{1}{4}r^4\sin^2 2\theta}$$
Then I take the limit:
$$\lim_{r\to0} r^{\frac{1}{4}r^4\sin^2 2\theta}$$
But since we have the $0^0$ indeterminacy, on the assumption that the limit is $L$, I proceed to take the natural logarithm of both sides:
\begin{align*}
\ln L = \ln\lim_{r\to0} r^{\frac{1}{4}r^4\sin^2 2\theta} &\stackrel{\text{nl is continuous}}{=} \lim_{r\to0} \ln (r^{\frac{1}{4}r^4\sin^2 2\theta}) \\
 &= \lim_{r\to0}\frac{1}{4}r^4\sin^2 2\theta \ln r \\
&= \lim_{r\to0}\frac{\sin^2 2\theta}{4} \lim_{r\to0}r^4\ln r
\end{align*}
I apply L'Hôpital's rule to the second limit:
$$\lim_{r\to0}r^4\ln r = \lim_{r\to0} \frac{\frac{1}{r}}{\frac{-4}{r^3}} = \lim_{r\to0} - \frac{r^2}{4} = 0$$
Since the other limit was bounded,
$$\ln L  = 0$$
Therefore the limit is $1$.
 A: Using polar form, you have successfully shown: assuming you know that $\lim\limits_{(x, y) \to (0, 0)}(x^2 + y^2)^{x^2y^2}$ exists, the limit has to be $1$. If you want to also show existence of the limit in the first place, you have to be a bit more careful.
Let us understand more precisely what you actually did when you converted the problem to polar form and showed that $\lim\limits_{r \to 0}r^{\frac{1}{4}r^4 \sin^2(2 \theta)} = 1$. The details of your work are correct but note that this limit does not involve the variable $\theta$ in anyway. Thus, what your work actually showed was that if the angle $\theta$ were kept fixed, then the expression $r^{\frac{1}{4}r^4 \sin^2(2 \theta)} \to 1$ if $r \to 0$. In other words, you showed that if you approach the origin over any fixed line, then the limit is $1$.
However, to show existence of the limit, you have to also allow the angle $\theta$ to vary arbitrarily while $r$ approaches $0$. That is, the limit has to be $1$ regardless of whether you approach the origin via a fixed line, a random curve, some combination of the two, etc. That said, here is a rearrangement of your work achieving exactly that:
So let us look at the expression $F(r, \theta) = r^{\frac{1}{4}r^4 \sin^2(2 \theta)}$. If $r \neq 0$, then the expression is positive. So, taking the natural logarithm of this expression is valid: $\ln(F(r, \theta)) = \frac{1}{4}r^4 \sin^2(2 \theta)\ln(r)$. I will now show that the absolute value of this expression approaches $0$:
\begin{align*}
0 \leq |\ln(F(r, \theta))| &= \Big|\frac{1}{4}r^4 \sin^2(2 \theta)\ln(r)\Big| \\
&= \Big|\frac{1}{4}r^4\ln(r)\Big|| \sin^2(2 \theta)| \\
&\leq \Big|\frac{1}{4}r^4\ln(r)\Big| \quad\text{ as $0 \leq \sin^2(2 \theta) \leq 1$ }
\end{align*} Essentially, by using the absolute value, I have gotten rid of the $\sin^2(2 \theta)$ term and now you can reason purely in terms of $r$. We can now make use of your work where you showed that $\frac{1}{4}r^4\ln(r) \to 0$ as $r \to 0$. Now, as you approach the origin regardless of how $\theta$ varies, $|\frac{1}{4}r^4\ln(r)|$ will go to $0$ by your work and the fact that it does not depend on $\theta$. As $|\ln(F(r, \theta))|$ is stuck between $|\frac{1}{4}r^4\ln(r)|$ and $0$, it must also go to $0$ (as $r \to 0$ and $\theta$ arbitrary) by the Squeeze Theorem, regardless of how $\theta$ varies. Well, if the absolute value of an expression approaches $0$, I guess the original expression must to do so too. So, $\lim\limits_{r \to 0,\ \theta \text{ arbitrary}}\ln(F(r, \theta) = 0$ and taking the exponent on both sides and using its continuity gives $\lim\limits_{r \to 0,\ \theta \text{ arbitrary}}F(r, \theta) = 1$ as desired.
See this thread Why exactly limit in polar coordinates isn't sufficient to find the limit in two variables? to further understand what things can go wrong if you do not reason with limits in polar form carefully. In particular, the answer in that thread gives you a way of showing both the existence of limits in polar form and calculating the limit's value (if it exists).
