# Proving that there is only one limit of a function using epsilon delta

Suppose $f:D \rightarrow R$ with $x_0$ as an accumulation point of D. Assume $L_1$ and $L_2$ are limits of $f$ at $x_0$. Prove $L_1=L_2$

If $x_0$ is an accumulation point of D, then $\exists y \ne x_0$ such that for some $\epsilon>0$, $(y- \epsilon,y + \epsilon) \in D$

If $\forall \epsilon>0$, $\exists \delta_1>0$ such that $\lvert f(x)-L_1 \rvert < \frac{\epsilon}{2}$, $\forall x$ with $0<\lvert x-x_0 \rvert <\delta_1$

and, $\forall \epsilon>0$, $\exists \delta_2>0$ such that $\lvert f(x)-L_2 \rvert < \frac{\epsilon}{2}$, $\forall x$ with $0<\lvert x-x_0 \rvert <\delta_2$

Let $\delta =$ min{$\delta_1,\delta_2$} $\forall x$ with $0<\lvert x-x_0 \rvert <\delta$

We want to find $\lvert L_1 -L_2 \rvert =0$ or $\lvert L_1-L_2 \rvert< \epsilon$ for some small value of $\epsilon>0$

$\lvert L_1-L_2 \rvert = \lvert L_1 + f(x) - f(x) - L_2 \rvert \le \lvert L_1-f(x) \rvert + \lvert f(x)-L_2 \rvert <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Hint: supposed that $L_1\neq L_2$ take $\displaystyle\epsilon=\frac{|L_1-L_2|}{2}$ and have a contradiction.
• Then $\lvert L_1-L_2 \rvert < \frac{\lvert L_1-L_2 \rvert}{4}+ \frac{\lvert L_1-L_2 \rvert}{4}= \frac{\lvert L_1-L_2 \rvert}{2}$ which causes a contradiction? – Ryan T. Donnelly Nov 7 '15 at 18:25
• @RyanT.Donnelly you have that $\lvert L_1-L_2 \rvert< \epsilon$, but $|L_1-L_2|>\epsilon=\frac{|L_1-L_2|}{2}$ this is the contradiction. – AsdrubalBeltran Nov 7 '15 at 18:42
No: "We want to find $\lvert L_1 -L_2 \rvert =0$ or $\lvert L_1-L_2 \rvert< \epsilon$ for some small value of $\epsilon>0$" is wrong. The condition $\lvert L_1-L_2 \rvert< \epsilon$ must hold for every $\epsilon>0$!
In addition, the sentence "If $x_0$ is an accumulation point of D, then $\exists y \ne x_0$ such that for some $\epsilon>0$, $(y- \epsilon,y + \epsilon) \in D$" is confused and essentially useless in this form.