Finding $\lim_{(x,y)\rightarrow (0,0)}\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$ I want to find $\lim_{(x,y)\rightarrow 0}\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$.
What I tried:
Denote $x=r\cos(\theta)$ and $y=r\sin(\theta)$. So the limit is:
$$\lim_{r\rightarrow 0}\frac{e^{-\frac{1}{r^2(\cos^2(\theta)+\sin^2(\theta))}}}{r^4(\cos^4(\theta)+\sin^4(\theta))} = \lim_{r\rightarrow 0}\frac{e^{-\frac{1}{r^2}}}{r^4(\cos^4(\theta)+\sin^4(\theta))}$$
Am I on the right track? If so, how do I continue?
 A: Now you can observe that $$ \frac{d}{d\theta}\left(\cos^{4}\left(\theta\right)+\sin^{4}\left(\theta\right)\right)=-\sin\left(4\theta\right)
 $$ and $-\sin\left(4\theta\right)=0
 $ if $\theta=\pi n/4,\,n \in \mathbb{Z}
 $. Hence we have $$\cos^{4}\left(\theta\right)+\sin^{4}\left(\theta\right)\geq\frac{1}{2}
  $$ thus $$0\leq\lim_{r\rightarrow0^{+}}\frac{e^{-1/r^{2}}}{r^{4}\left(\cos^{4}\left(\theta\right)+\sin^{4}\left(\theta\right)\right)}\leq2\lim_{r\rightarrow0^{+}}\frac{e^{-1/r^{2}}}{r^{4}}=0.
 $$ To prove the last limit note that $$2\lim_{r\rightarrow0^{+}}\frac{e^{-1/r^{2}}}{r^{4}}=2\lim_{y\rightarrow\infty}\frac{y^{4}}{e^{y^{2}}}\overset{H}{=}4\lim_{y\rightarrow\infty}\frac{y^{2}}{e^{y^{2}}}\overset{H}{=}4\lim_{y\rightarrow\infty}\frac{1}{e^{y^{2}}}=0
 $$ where $H
 $ indicates that I used De L'Hopital's rule and $y=1/r$.
A: Let
$1/r^2
= z
$.
Then
$U(r, \theta)
=\frac{e^{-\frac{1}{r^2}}}{r^4(\cos^4(\theta)+\sin^4(\theta))}
=\frac{z^2e^{-z}}{\cos^4(\theta)+\sin^4(\theta)}
$.
Also,
$\begin{array}\\
\cos^4(\theta)+\sin^4(\theta)
&=\cos^4(\theta)+\sin^4(\theta)+2\cos^22(\theta)\sin^2(\theta)-2\cos^2(\theta)\sin^2(\theta)\\
&=(\cos^2(\theta)+\sin^2(\theta))^2-2\cos^2(\theta)\sin^2(\theta)\\
&=1-\frac12(2\cos(\theta)\sin(\theta))^2\\
&=1-\frac12(\sin(2\theta))^2\\
&\ge 1-\frac12\\
&=\frac12
\end{array}
$
so
$|U(r, \theta)|
\le\frac12 z^2e^{-z}
$.
Since
$\lim_{z \to \infty} z^2e^{-z}
= 0
$,
as has been shown here many times,
$\lim_{r \to 0}|U(r, \theta)|
= 0
$.
(Added later)
To show
$\lim_{z \to \infty} z^2e^{-z}
= 0
$,
from the power series
for $e^z$,
$e^z \ge \frac{z^3}{6}
$,
so
$\frac{z^2}{e^z}
\le \frac{6}{z}
\to 0
$.
Note that this easily generalizes:
To show
$\lim_{z \to \infty} z^ne^{-z}
= 0
$,
from the power series
for $e^z$,
$e^z \ge \frac{z^{n+1}}{(n+1)!}
$,
so
$\frac{z^n}{e^z}
\le \frac{(n+1)!}{z}
\to 0
$.
