Prove that limits from the left and right $\lim_{x\to\ a^+} f(x)= \lim_{x\to\ a^-}=L$ then $\lim_{x\to\ a} f(x)=L$. Prove that $\lim_{x\to\ a^+} f(x)= \lim_{x\to\ a^-}=L$ then $\lim_{x\to\ a} f(x)=L$.
I started by assuming that there exists some interval $(c,a)$ such that any sequence $x_n$ in $(c,a)$ converging to $a$ has $\lim_{x\to\ a}f(x_n)$ and also some interval $(a,b)$ such that any sequence $x_n$ in $(a,b)$ converging to $a$ has $\lim_{x\to\ a}f(x_n)$.
I’m not really sure where to go from here. I want to define a sequence that spans the union of these two intervals, and then divide it into two subsequences that are less than and greater than $a$. Is that right? Any suggestions for proceeding?
 A: We might as well just get dirty with the $\epsilon-\delta$ approach. :)
As with any proof, we start with what knowledge the assumptions give us and see where we need to go. It should be clear where we need to go given the definition of the limit. So let $\epsilon>0$.
The assumptions then tell us that, for $x\to a^+$, there is an $\eta>0$ such that for all $x$,
$$0<x-a<\eta \implies |f(x) - L|<\epsilon$$
and that, for $x\to a^-$, there is a $\gamma>0$ such that for all $x$,
$$0<a-x<\gamma \implies |f(x) - L|<\epsilon$$
The definition of the limit tells us we need to find a $\delta>0$ such that for all $x$,
$$0<|x-a|<\delta \implies |f(x) - L|<\epsilon$$
Like most $\epsilon-\delta$ proofs, we need to make a clever choice regarding our $\delta$. I recommend sketching what approaching $a$ from the left and right looks like on the $x$-axis given $\gamma$ and $\eta$ as above. Coming from the right, we know that once we can get $x$ less than $a+\eta$ we have $|f(x)-L|<\epsilon$. Coming from the left, we know that once we can get $x$ greater than $a-\gamma$ we have $|f(x)-L|<\epsilon$. So we need a $\delta$ such that once $x$ is within either $a-\delta$ or $a+\delta$ we have $|f(x)-L|$.
If you sketch this situation out then it will become clear that you should let $\delta$ be smaller than both $\gamma$ and $\eta$. This squeezes $x$ in close enough to satisfy both conditions as described above. One choice is to let $\delta = \min\{\gamma, \eta\}$, that is let $\delta$ be equal to the smaller of the two numbers, $\gamma$ and $\eta$. Then $\delta\le\gamma$ and $\delta\le\eta$.
From there, I will let you work out the details in showing that this leads to satisfying both $0<x-a<\eta$ and $0<a-x<\gamma$, which will then imply $|f(x)-L|<\epsilon$ and thus $\lim_{x\to a} f(x) = L$.
