There are various ways to explain this, but probably the best way to start is to try to think of "continuous at a point" or "limit at a point" as being its own independent concept, rather than something defined in terms of right-continuity and left-continuity. In the case of the real numbers, there are a lot of ways to define this, but here are two good ones:
Let $D\subseteq \Bbb R$ and let $f\colon D\to \Bbb R$. Then $f$ is continuous at $c\in D$ iff:
For every $\epsilon > 0$ there is a $\delta >0$ such that for all $x\in D$ such that $|x-c|<\delta$, $|f(x)-f(c)| < \epsilon$.
Whenever $(x_i)$ is a sequence in $D$ that converges to $c$, $(f(x_i))$ converges to $f(c)$.
Edit:
Thus the notion of continuity at an endpoint is perfectly sensible. Also, it's quite reasonable to consider a derivative at an endpoint. I think you may be getting a little confused because there are so many theorems out there that apply when a function is continuous on an interval and differentiable in its interior—it's not that the function can't be differentiable there, but that the theorem doesn't need it to be.
Yes.
The keystone is:
Lemma. Let $f\colon [a,b]\to\mathbb R$ be continuous and assume that $f'_+(x)$ exists and is $>0$ for all $x\in [a,b)$. Then $f$ is strictly increasing.
Assume otherwise, i.e. $f(a)\ge f(b)$.
We recursively define a map $g\colon \operatorname{Ord}\to [a,b)$ such that $g$ and $f\circ g$ are strictly inreasing. Since the class $\operatorname{Ord}$ of ordinals is a proper class and $g$ is injective, we arrive at a contradiction, thus showing the claim.
- Let $g(0)=a$.
- For a successor $\alpha=\beta+1$ assume we have already defined $g(\beta)$. For sufficently small positive $h$ we have that $g(\beta)<g(\beta)+h<b$ and $\frac{f(g(\beta)+h)-f(g(\beta))}{h}\approx f_+(g(\beta))>0$. Pick one such $h$ and let $g(\alpha)=g(\beta)+h$.
- If $\alpha$ is a limit ordinal, assume $g(\beta)$ is defined for all $\beta<\alpha$. Let $x=\sup_{\beta<\alpha} g(\beta)$. A priori only $x\le b$, but we need $x<b$. Because $f$ is continuous and $f\circ g$ is strictly increasing, we conclude that $f(x)=\sup_{\beta<\alpha} f(g(\beta))\ge f(g(1))>f(g(0))=f(a)=f(b)$. Therefore $x<b$ as desired and we can let $g(\alpha)=x$.
$\square$
Corollary 1. (something like a one-sided Rolle theorem) Let $f\colon [a,b]\to\mathbb R$ be continuous with $f(a)=f(b)$. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=0$ for some $x\in[a,b)$.
Proof. Assume otherwise. Then either $f_+(x)>0$ for all $x$ or $f_+(x)<0$ for all $x$. In the first case the lemma applies and gives us a contradiction to $f(a)=f(b)$; in the other case, we consider $-f$ instead of $f$. $\square$
Corollary 2. (something like a one-sided IVT) Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in[a,b)$.
Proof. Apply the previous corollary to $f(x)-\frac{f(b)-f(a)}{b-a}x$. $\square$
By symmetry, we have
Corollary 3. Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_-$ exists and is continuos in $(a,b]$. Then $f'_-(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in(a,b]$. $\square$
Theorem. Let $f\in C(\mathbb R)$ be a function with $f'_-$ continuous on $\mathbb R$.
Then $f\in C^1(\mathbb R)$.
Proof.
Consider aribtrary $a\in \mathbb R$.
Let $\epsilon>0$ be given.
Then by continuity of $f'_-$, for some $\delta>0$ we have $|f'_-(x)-f'_-(a)|<\epsilon$ for all $x\in(a,a+\delta)$.
Thus for $0<h<\delta$ we have $\left|\frac{f(a+h)-f(a)}{h}-f'_-(a)\right|<\epsilon$ by corollary 3. We conclude that $f'_+(a)=f'_-(a)$, i.e. $f$ is differentiable at $a$. $\square$
Best Answer
Consider the function:
g(x) := \begin{array}{rl} x^2 \sin(1/x), & x \ne 0 \\ 0, & x = 0 \\\end{array}.
Then, we have:
g'(x) = \begin{array}{rl} 2x \sin(1/x) - \cos(1/x) & x \ne 0 \\ 0, & x = 0 \\\end{array}
Observe that $g'$ is not continuous at $0$ but $g'(0)=0$.