Lateral limits of an endpoint of the interval. Imagine we have a the domain $D=[d_1,d_2]$ of a continuous function $f$.
The definition of right limit I'm using is the following:
$$\lim_{x\rightarrow a^+}f(x)=b \Leftrightarrow \forall_{\epsilon}\exists_{\delta}\forall_{x}(x\in D \ \cap \ ]a,a+\delta[\ \Rightarrow \ f(x) \in N_{\epsilon}(b) )$$, where $N_{\epsilon}(b)$ is the neighbourhood of length $2\epsilon$ at point $b$. We define similarly a left limit.
If I pick  point $a=d_2$, which is a limit point(<=> accumulation point), then the implication is vacuously true, for any value $b$...
Then how can I say that there's no $\displaystyle \lim_{x\rightarrow d_2^+}f(x)$? 
Or when I say that $\lim_{x\rightarrow d_2^-}f(x)=\lim_{x\rightarrow d_2^+}f(x) \Leftrightarrow \lim_{x\rightarrow d_2}f(x) \text{ exists }$, it's valid only for interior points?
Thanks.

the motivation of limit is to describe the 'behaviour' of a function, regradless the the value of the function at the point of consideration. For example, we have
\begin{eqnarray*}
f(x)&=&x+1, \; x\neq 0\\
    &=&-1, \; x=0.
\end{eqnarray*}
Then we have $\lim_{x\rightarrow 0}f(x)=1$ even though $f(0)=-1$, because around $x=0$, the 'behaviour' of the fanction is tends to $1$. That is why, the point of consideration (the point you want to evaluate the limit) has to be an accumulation point, because by doing that we are sure we have 'enough' point around the point of consideration to observe the 'behaviour' of the function.
For your question, because we are discussing about right limit, then we have to observe the 'behaviour' of the function from the right. That is why, the point of consideration has to be an accumulation point from the right. Since $d_2$ is the maximal point in $[d_1,d_2]$ and the function is undefined when $x>d_2$, then $d_2$ is not an accumulation point from the right, and we cannot observe the 'behaviour' of $f$ from the right. I hope it is clear enough, let me know if you still doubt about something.
