Please, how find the limit of
$$\lim \limits_{(x,y) \to (0,0)}\frac{e^{-1/(x^2+y^2)}}{x^4+y^4}$$
So i tried to substitute t
$$\lim \limits_{t \to 0^+}\frac{e^{-1/t}}{t^2}$$
I substituted a=1/t
$$\lim \limits_{a \to \infty}\frac{a^2}{e^a}=0$$
Before asking, I tried using polar coordinates
$$\lim \limits_{r \to 0}\frac{e^{-1/r^2}}{r^4(sin^4\theta+cos^4\theta)}$$
We have that
$$\frac{e^{-1/(x^2+y^2)}}{x^4+y^4}=\frac{e^{-1/(x^2+y^2)}}{(x^2+y^2)^2}\frac{(x^2+y^2)^2}{x^4+y^4} \to 0$$
indeed for $t=x^2+y^2 \to 0^+$ $$\frac{e^{-\frac1t}}{t^2}\to 0$$
and by polar coordinates
$$0\le \frac{(x^2+y^2)^2}{x^4+y^4}=\frac{1}{\cos^4 \theta+\sin^4 \theta}\le M\in\mathbb{R}$$
which is indeed bounded
Polar coordinates are also a good approach here, but maybe a bit more work than necessary.
Note that $$\sin^4t+\cos^4t = (\sin^2t + \cos^2t)^2-2\sin^2t\cos^2 t = 1 - \frac 12 \sin^2(2t) \\ \implies \frac 12\leq \sin^4t+\cos^4t\leq 1$$ and thus
$$\frac{e^{-1/r^2}}{r^4(\sin^4\theta+\cos^4\theta)} \leq \frac{2}{r^4e^{1/r^2}}.$$
Now, since $r^5e^{1/r^2}\to \infty$ as $r\to 0^+$, there is $\delta'>0$ such that $$0<r<\delta' \implies r^5e^{1/r^2} > 2 \implies \frac{2}{r^4e^{1/r^2}} < r.$$
Finally, choose $\delta = \min\{\delta',\varepsilon\}$ to conclude that $$0<\sqrt{x^2+y^2}<\delta \implies 0 < r < \delta \implies |f(x,y)|<\varepsilon.$$
