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Let $X \subset \mathbb{R}, \; f: X \rightarrow \mathbb{R}$ be a function and $ a \in X'$. If $\lim_{x \rightarrow a}f(x)= L_1$ and $\lim_{x \rightarrow a}f(x)= L_2$ then $L_1 = L_2$.



Given $\varepsilon >0$, there exists $\delta_1 >0$ and $\delta_2 >0$, such that, $x \in X$,$0 < |x-a| < \delta_1 \Longrightarrow |f(x)- L_1| < \varepsilon /2 $ and $0 < |x-a| < \delta_2 \Longrightarrow |f(x)- L_2| < \varepsilon /2 $. Let $\delta = \min \{\delta_1,\delta_2 \}$. $a \in X'$ and we can obtain $x' \in X$, such that $0 < |x'-a| < \delta$ then $|L_1 - L_2| \leq |L_1 - f(x') | + |f(x') - L_2| < \varepsilon/2 + \varepsilon /2 = \varepsilon$. There for $|L_1 - L_2 | < \varepsilon$ for every $\varepsilon >0$, this implies $L_1 = L_2$