# 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

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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.

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What is the isolated point in my example supposed to be? It can't be $-6$. Having trouble relating the two examples – Hawk Dec 24 '12 at 22:02
The example I gave is not a case of what you asked about. Sorry if that was unclear. In your example there exists one limit, the right-hand one. What I was trying to demonstrate is that it is possible for a function to be continuous at a point without having a left or right-hand limit. So the presence of either limit is not strictly necessary for continuity, so you have to look at the definition. – user54147 Dec 24 '12 at 22:52
I am getting caught up with the reason that since left limit doesn't exist (and hence not equal to the right), the real limit wouldn't exist and therefore the function wouldn't be continuous – Hawk Dec 24 '12 at 22:55
The problem is that isn't exactly how continuity is defined. The notion of left and right limits is not necessary or sufficient, for defining continuity. Using $\epsilon - \delta$ definitions, or sequences, will make the fact clearer. As I said before, if a function is defined on an isolated point, clearly no limits exist at all, yet the function is continuous there. In "Principles of Mathematical Analysis", Rudin only defines right and left limits for functions defined on open intervals, otherwise, you have to use the proper definitions. – user54147 Dec 25 '12 at 3:10

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.

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But my point is that it doesn't even make sense to define a left limit, so we can't say the right and left agree, and therefore the limit shouldn't exist (and hence not continuous). – Hawk Dec 24 '12 at 20:18
@sizz Read the statement I said is not correct. – Nameless Dec 24 '12 at 20:19