Why does this limit exist and this function continuous? Consider $f(x) = x\sqrt{x+6}$ on $[-6,0]$
Apparently this function the limit $x \to -6$ exists and we can conclude it is continuous at $x = -6$. 
However, what i cannot comprehend is that the left limits actually exist, 
Wolfram Plot of $f$
For small $h$ just outside $(-\infty,-6)$, isn't the function undefined (the real valued function)? So I simply cannot comprehend how the limit exists here
 A: In this case $f$ is a function $f:[-6,0] \to \mathbb{R}$, as you said. So the points to the left of $x=-6$ are irrelevant, for our purposes they don't exist. Then, by the definition of continuity at $x=-6$ we are only concerned with showing that $|f(-6)-f(x)|< \epsilon$ when $x \in [-6,0]$ for any given $\epsilon$, given that $|-6-x|<\delta$ for some $\delta$.
We can have an even stricter example: if $E \subset \mathbb{R}$ and $x$ is an isolated point of $E$, and $f$ is defined at $x$, then $f$ is necessarily continuous at $x$: just think about how it is then trivial to find $\delta$. Since $f$ isn't defined anywhere right next to $x$, for a sufficiently small $\delta$-neighbourhood of $x$, $f(x)$ will be the only value that $f$ can take in that neighbourhood, so clearly $|f(x)-f(t)|=|f(x)-f(x)|=0<\epsilon$ as long as $|x-t|< \delta$.
In this example I gave there are no left-hand OR right-hand limits, since it is an isolated point, yet the function is continuous there.
A: Firstly continuity at end-points is traditionally defined via one sided limits.
$f$ is continuous at $-6$ in the sense that $\lim_{x\to 6^+}f(x)$ exists and is $f(6)$
More formally: $f$ is continuous in $[a,b]$ if it is continuous in its interior and $\lim_{x\to a^+}f(x)=f(a)$ and $\lim_{x\to b^-}f(x)=f(b)$
Secondly the limit $\lim_{x\to a^-}f(x)$ may not be defined (in the sense that $x$ can't approach $a$ from the left) but the limit $\lim_{x\to a}f(x)$ does exist. In fact: 
If $f:X\to \mathbb{R}$ and $a$ is an accumulation point of $X$ only from the right (*) and $\lim_{x\to a^+}f(x)=L$ then $\lim_{x\to a}f(x)=L$
(*): That is $\forall \delta>0$, $(a,a+\delta)\cap X\neq \emptyset$ (accumation from the right) while $\exists\delta>0$, $(a-\delta,a)\cap X=\emptyset$  (non accumulation from the left)
Note: The statement: the limit exists iff the one sided lmits exist and are equal is not correct. The correct statement is the following: 
The limit exists at point that is a limit point from the left and from the right iff the one sided lmits exist and are equal
The limit exists at point that is a limit point only from the left iff the left one sided lmits exists.
The limit exists at point that is a limit point only from the right iff the right one sided lmits exists.
