Compute limit of $f(x,y)=(x^2+y^2)^{\left|x\right|}$ $$f(x,y)=(x^2+y^2)^{\left|x\right|} $$ find limit at $(0,0)$
moving into parametric form $f(r\cos\phi,r\sin\phi)=(r^2)^{r\left|\cos\phi\right|}= e^{2r\left|\cos\phi\right|\ln r}$
$\bigl|r\left|\cos\phi\right|\ln r\bigr|\leq r\times \ln(\frac{1}{r})$ which goes to $0$
so $$f(x,y) \leq e^{2\sqrt{x^2+y^2}\ln(\frac{1}{\sqrt{x^2+y^2}})}$$ which goes to $1$ as $(x,y) \to (0,0)$
I am stuck how conclude that limit is actually $1$? I get that $f(x,y)$ is less than some function which goes to $1$.
 A: We get,
\begin{align*}
\lim_{(x,y)\to (0,0)} (x^{2}+y^{2})^{|x|}&=\lim_{r\searrow 0}(r^{2})^{|r\cos \theta|}, \tag{1}\\
&=\lim_{r\searrow 0} r^{|2r\cos \theta|}, \\
&=\lim_{r\searrow 0} e^{\log r^{|2r\cos \theta|}}, \tag{2}\\
&=e^{\displaystyle \lim_{r\searrow 0} \log r^{|2r\cos \theta|}} , \\
&=e^{\displaystyle\lim_{r\searrow 0}|2r\cos \theta|\log r}, \tag{3}\\\
&=e^{0}, \\
&=1.
\end{align*}
where $(1)$ is change of variables to polar coordinates $x=r\cos \theta, y=r\sin \theta$ with $r>0$ and $\theta\in [0,2\pi[$ and $(2)$ is just $x^{y}=e^{\log x^{y}}$. Finally, $(3)$ is given because $\displaystyle\lim_{r\searrow 0}|2r\cos \theta|\log r=0$ because it is bounded. Hence,
$$\boxed{\lim_{(x,y)\to (0,0)}(x^{2}+y^{2})^{|x|}=1}$$

More details:
Notice that,
$$-1\leqslant \cos \theta \leqslant 1,$$ $$-|2r|\log r\leqslant |2r|\cos \theta \leqslant |2r|\log r,$$ $$0\leqslant \lim_{r\searrow 0}|2r|\cos\theta \leqslant 0.$$
Hence, $$\lim_{r\searrow 0} |2r\cos \theta|\log r=0.$$
A: Note that $\lim_{x \to 0} x \ln(x) = \lim_{x \to 0} \frac{\ln(x)}{\frac{1}{x}} = \lim_{x \to 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = 0$.
Now as $e^x$ is continuous and increasing over the reals, we have
$$e^{0} \leq e^{2r |\cos \theta| \ln(r)} \leq e^{2r \ln r}$$
\begin{align*}
&\implies \lim_{r \to 0} e^{0} \leq \lim_{r \to 0} e^{2r |\cos \theta| \ln(r)} \leq \lim_{r \to 0} e^{2r \ln r} \\
&\implies 1 \leq \lim_{(x,y) \to (0,0)} f(x,y) \leq e^0  = 1\text{ (from the above  result)}
\end{align*}
By the Sandwich theorem, you have the result.
