Gerek Kısmı: $\lim\limits_{x\to a}f(x)=L$ ve $\epsilon>0$ olsun.
$\left.\begin{array}{rr} \epsilon>0 \\ \\ \lim\limits_{x\to a}f(x)=L \end{array}\right\}\Rightarrow (\exists\delta>0)(\forall x\in A)(0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon)$
$\Rightarrow (\exists\delta>0)\forall x(x\in A\Rightarrow [x\in (a-\delta,a)\cup (a,a+\delta)\Rightarrow f(x)\in (L-\epsilon,L+\epsilon)])$
$\Rightarrow (\exists\delta>0)\forall x(x\in A\cap [(a-\delta,a)\cup (a,a+\delta)]\Rightarrow x\in f^{-1}[(L-\epsilon,L+\epsilon)])$
$\Rightarrow (\exists\delta>0)( A\cap [(a-\delta,a)\cup (a,a+\delta)]\subseteq f^{-1}[(L-\epsilon,L+\epsilon)])$
$\Rightarrow (\exists\delta>0)(f[A\cap [(a-\delta,a)\cup (a,a+\delta)]]\subseteq (L-\epsilon,L+\epsilon))$
$\Rightarrow (\exists\delta>0)(f[A\cap (a-\delta,a)]\subseteq f[A\cap [(a-\delta,a)\cup (a,a+\delta)]]\subseteq (L-\epsilon,L+\epsilon))$
$\Rightarrow (\exists\delta>0)(f[A\cap (a-\delta,a)]\subseteq (L-\epsilon,L+\epsilon))$
O halde $$\lim\limits_{x\to a^-}f(x)=L$$ olur. Ayrıca $$\lim\limits_{x\to a^+}f(x)=L$$ olduğu da benzer şekilde gösterilir.
Yeter Kısmı: $\lim\limits_{x\to a^-}f(x)=\lim\limits_{x\to a^+}f(x)=L$ ve $\epsilon>0$ olsun.
$\left.\begin{array}{rr} \epsilon>0 \\ \\ \lim\limits_{x\to a^-}f(x)=L \end{array}\right\}\Rightarrow (\exists\delta_1>0)(f[A\cap (a-\delta_1,a)]\subseteq (L-\epsilon,L+\epsilon))\ldots (1)$
$\left.\begin{array}{rr} \epsilon>0 \\ \\ \lim\limits_{x\to a^+}f(x)=L \end{array}\right\}\Rightarrow (\exists\delta_2>0)(f[A\cap (a,a+\delta_2)]\subseteq (L-\epsilon,L+\epsilon))\ldots (2)$
$\delta:=\min\{\delta_1,\delta_2\}\ldots (3)$
$\left.\begin{array}{rr}(1),(2),(3)\Rightarrow (\exists\delta>0)(f[A\cap (a-\delta,a)]\cup f[A\cap (a,a+\delta)] \subseteq (L-\epsilon,L+\epsilon)) \\ \\ f[A\cap (a-\delta,a)]\cup f[A\cap (a,a+\delta)]=f[A\cap [(a-\delta,a)\cup (a,a+\delta)]]\end{array}\right\}\Rightarrow$
$\Rightarrow (\exists\delta>0)(f[A\cap [(a-\delta,a)\cup (a,a+\delta)]] \subseteq (L-\epsilon,L+\epsilon)).$