Solutions

Suppose

$$\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=l.$$

Pick any $\varepsilon>0$. By definition, there exist $\delta_1,\delta_2$ such that, for all $x$,

$$0<x-a<\delta_1 \Rightarrow |f(x)-l|<\varepsilon,$$

and

$$0<a-x<\delta_2 \Rightarrow |f(x)-l|<\varepsilon.$$

Let $\delta=\min(\delta_1,\delta_2)$. Combining both implications above, we have

$$0<|x-a|<\delta \Rightarrow |f(x)-l|<\varepsilon.$$