because KaTeX was being SPACE-SENSITIVE like who asked for that BS??
$f_{xy}=f_{yx}$
$E=mc^2 + AI$
\(f (x)\)
$e^{i \square}=\cos(\square)+i \sin(\square)$
$$\frac{\partial f}{\partial x} \neq \frac{\partial f}{\partial y}$$
and in calculus 1, we learn that
$$\forall \epsilon >0 , \exists \delta ~\text{S.T.} ~0<|x-a| <\delta \implies |f(x)-L| < \epsilon $$
$$a b$$
$$a b$$
$$a b$$
\[ non-space sensitive gotta use tildes \]
$$ \lim_{ (x,y) \to (a,b) }~f(x,y) =L \stackrel{\text{def}}{\iff} \forall \epsilon >0, \exists \delta >0 ~\text{S.T.}~ \sqrt{ (x-a)^2 +(y-b)^2 } < \delta \implies |f(x,y)-L| < \epsilon $$
$$ \lim_{ (x,y) \to (a,b) }~f(x,y) =L $$
$$\stackrel{\text{def}}{\iff} $$
$$\forall \epsilon >0, \exists \delta >0 ~\text{S.T.}~ \sqrt{ (x-a)^2 +(y-b)^2 } < \delta \implies |f(x,y)-L| < \epsilon $$
$$\begin{aligned}&\lim_{ (x,y) \to (a,b) }~f(x,y) =L \\\\&{\Updownarrow}\\ \\ &\forall\epsilon >0, \exists \delta >0 ~\text{S.T.}~ \sqrt{ (x-a)^2 +(y-b)^2 } < \delta \implies|f(x,y)-L| < \epsilon \end{aligned}$$
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