How to formally justify that the limit as $(x,y) \to (0,0)$ of the function $e^{x^{2} + y^{2}}$ exists? As my question asks: how to formally justify that the limit as $(x,y) \to (0,0)$ of the function $e^{x^{2} + y^{2}}$ exists?
It is clear by "plugging in" that I will get a limit of $1$, that is $$\lim_{(x,y) \to (0,0)} e^{x^{2} + y^{2}} = 1$$ but that is not a formal justification.
One thing I thought of was using the following:

*

*it has been shown that $$\lim_{(x,y) \to (0,0)} x^{2} + y^{2} = 0$$


*from one variable calculus we have: $$\lim_{t \to 0}e^{t} = 1$$
If we consider the composition $g(f(x,y))$ where $f(x,y) = x^{2} + y^{2}$ and $g(t) = e^{t}$, the  result will follow. Is this enough to be consider a "formal" justification?
 A: This idea can be formalized using the analysis of metric spaces. Consider the metric space $M_1 =(\mathbb{R}^2,d_1)$ where $d_1:\mathbb{R}^2\times \mathbb{R}^2 \to \mathbb{R}\: \:$ and
$$ d_1(x,y) = \sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2} $$
for all $x,y\in\mathbb{R}^2$.
Also consider the metric space $M_2 =(\mathbb{R},d_2)$ where $d_2:\mathbb{R}\times \mathbb{R} \to \mathbb{R}\: \:$ and
$$ d_2(x,y) = | x - y | $$
for all $x,y\in\mathbb{R}$.
Pf:
Let $\varepsilon > 0$ and let $\delta = \sqrt{\ln{(1+\varepsilon)}}$. Observe that if an element $(x,y) = p\in \mathbb{R}^2$ is in a neighborhood of the ordered pair $(0,0) = \vec{0}$ and satisfies
$$ d_1(p,\vec{0}) < \delta $$
then $$\begin{align}& &\sqrt{x^2+y^2}&<\sqrt{\ln{(1+\varepsilon)}} \\\\
&\implies &x^2+y^2&<\ln{(1+\varepsilon)} \\\\
&\implies &e^{x^2+y^2}&<(1+\varepsilon) \\\\
&\implies &e^{x^2+y^2} - 1 &<\varepsilon \\\\
&\implies &|e^{x^2+y^2} - 1| &<\varepsilon  \\\\
&\implies &d_2(e^{x^2 + y^2},1) &<\varepsilon \text{ .} \\
\end{align}$$
From this, we can conclude by the definition of the limit that a function $f:M_1\to M_2$ given as $f(x,y) = e^{x^2 + y^2}$ has the limit
$$ \lim_{(x,y)\to (0,0)} f(x,y) = 1 $$
$$\tag*{$\blacksquare$}$$
A: Let me, additionally to well know result about composition of continuous functions, bring one result which justifies existence of limit for not continuous case:
If exists, finite or infinite, limits $\lim\limits_{x \to a}f(x)=b$ and $\lim\limits_{y \to b}F(x)$, and in some punctured neighbourhood of $a$ we have $f(x)\ne b$, then exists limit of composition and holds
$$\lim\limits_{x \to a}F(f(x)) = \lim\limits_{y \to b}F(x)$$
Theorem is true in multidimensional case also.
Now to use this theorem in our case is enough to note that $f(x,y)=x^2+y^2 \ne 0$ is equivalent $(x,y)\ne (0,0)$.
