Existence of a limit via one-sided limits

($\Rightarrow$) Suppose $\lim_{x\to x_0} f(x) = L$. Given $\varepsilon>0$, the ε–δ definition gives $\delta>0$ such that $0<|x-x_0|<\delta\Rightarrow |f(x)-L|<\varepsilon$.

In particular, if $x_0<x<x_0+\delta$ (the right-hand condition) or $x_0-\delta<x<x_0$ (the left-hand condition), we automatically have $0<|x-x_0|<\delta$, so $|f(x)-L|<\varepsilon$. Thus both one-sided limits equal $L$.

($\Leftarrow$) Conversely, suppose both one-sided limits equal $L$. For a given $\varepsilon>0$, the left-hand limit gives $\delta_1>0$ working on $(x_0-\delta_1,x_0)$, and the right-hand limit gives $\delta_2>0$ working on $(x_0,x_0+\delta_2)$.

Set $\delta=\min(\delta_1,\delta_2)>0$. Then any $x$ with $0<|x-x_0|<\delta$ lies either in $(x_0-\delta_1,x_0)$ or in $(x_0,x_0+\delta_2)$, and in both cases $|f(x)-L|<\varepsilon$. Hence $\lim_{x\to x_0} f(x) = L$. $\blacksquare$