Solutions

Suppose that

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

Let

$$\varepsilon=\frac{m-l}{2}.$$

By definition, there exist $\delta_1,\delta_2$ such that, for all $x$,

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

and

$$0<y-a<\delta_2 \Rightarrow |f(y)-m|<\varepsilon.$$

Let $\delta=\min(\delta_1,\delta_2)$. When

$$x<a<y\quad \wedge\quad a-x<\delta\quad \wedge\quad y-a<\delta,$$

we have

$$x<a<y\quad \wedge\quad |x-a|<\delta\quad \wedge\quad |y-a|<\delta.$$

Then

$$f(x)<\frac{l+m}{2}<f(y).$$

Also the converse is not necessarily true. We could have

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