The rigorous definition of limit for the function of two variable is given as following:
Give a function of two variables $f : D → \mathbb{R}, D ⊆ \mathbb{R}^2$ such that $D$ contains points arbitrarily close to a point $(a, b)$, we say that the limit of $f (x, y)$ as $(x, y)$ approaches $(a, b)$ exists and has value $L$, if and only if for every real number $ε > 0$ there exists a real number $δ > 0$ such that
$$|f (x, y) − L| < ε$$
whenever $$ 0 <\sqrt{(x − a)^2 + (y − b)^2}< δ.$$ we then write $$ \lim_{(x,y)\to(a,b)} f(x,y)=L.$$
I want to apply this definition to the function $ f(x,y)=e^{-x^2y},$ but unlike the single variable cases, I could not find a way to substitute the $δ$ in to the inequality, and then choose a minimum value to show that the $ε$ is bounded. Could some one help me on this problem? Thanks.
Warning: Using only first principles, this is extremely tedious.
Let $(x_0,y_0)$ be given and fixed. First choose $\epsilon>0$ sufficiently small ($\epsilon<e^{-x_0^2y_0}$) so that \begin{align} 0<e^{-x_0^2y_0}-\epsilon<e^{-x^2y}<e^{-x_0^2y_0}+\epsilon \\ \implies \ln(e^{-x_0^2y_0}-\epsilon)<-x^2y<\ln(e^{-x_0^2y_0}+\epsilon) \\ \implies \color{red}{\ln(e^{-x_0^2y_0}-\epsilon)+x_0^2 y_0}<\color{blue}{-x^2y+x_0^2 y_0}<\color{green}{\ln(e^{-x_0^2y_0}+\epsilon)+x_0^2 y_0}. \end{align} Note that the red is negative and the green is positive. By adding and subtracting $x^2y_0$ respectively, the middle term can be written as \begin{align} (x_0^2-x^2)y_0+x^2(y_0-y)=(x_0-x)(x_0+x)y_0+x^2(y_0-y). \label{*}\tag{*} \end{align} First start with $\delta_1=1$. Then $-1<x-x_0<1\implies -1+2x_0<x+x_0<1+2x_0$ so $|x+x_0|\leq C_1(x_0):=\max\{|-1+2x_0|,|1+2x_0|\}$.
Similarly, $-1<x-x_0<1\implies x_0-1<x<x_0+1\implies |x|<C_2(x_0):=\max\{|x_0-1|,|x_0+1|\}.$ Therefore, \eqref{*} is bounded above by $$ |x_0-x||x_0+x||y_0|+|x|^2|y_0-y|\leq C_1|y_0|\delta_2+C_2^2\delta_2=(C_1|y_0|+C_2^2)\delta_2 $$ with $\delta_2$ to be chosen. Choose $\delta_2=\frac{\ln(e^{-x_0^2y_0}+\epsilon)+x_0^2 y_0}{C_1|y_0|+C_2^2}$. This part takes care of showing the blue is less than the green above.
Similar techniques show \eqref{*} is bounded below by $$ -|x_0-x||x_0+x||y_0|-|x|^2|y-y_0|\geq -(C_1|y_0|+C_2^2)\delta_3. $$ Choose $\delta_3=-\frac{\ln(e^{-x_0^2y_0}-\epsilon)+x_0^2 y_0}{C_1|y_0|+C_2^2}$. This will take care of showing that the blue is greater than the red.
In conclusion, let $\delta=\min\{\delta_1,\delta_2,\delta_3\}$. Then when $\sqrt{(x-x_0^2)+(y-y_0)^2}<\delta$, we will have red is less than blue is less than green. Working backwards shows $|e^{-x^2y}-e^{-x_0^2 y_0}|<\epsilon$.
If $\epsilon$ is not small, find $\epsilon_1<\epsilon$ that is and show everything is less than $\epsilon_1$ (using above). Then everything will be less than $\epsilon$.