$\lim_{x \to 4} \sqrt{x^2-16}$ Not clear to me why the limit as $x$ goes to $4$ of $\sqrt{x^2-16}$ is $0$, since the limits on both sides of $4$ are not the same. From the right it is zero, but from the left ($x= 3.99999$) is undefined.  
 A: We can assert $\lim_{x \to 4} \sqrt{x^2-16} = 0$ unambiguously. Applying the definition of the limit to this, we are saying
$\forall \epsilon>0 \ \exists \delta>0$ such that $\forall x \in \text{dom}{(f)}$ satisfying $0<|x-4|<\delta$, $|\sqrt{x^2-16}|<\epsilon$. 
In a small enough neighbourhood of $4$, there exist no $x$ satisfying the condition $x \in \text{dom}{(f)}$ to the left of $4$. We thus need only consider the right-hand side. The function is right-continuous at $x=4$ and so the condition is met. 
If we didn't have this definition, there would be some not-so-nice consequences; one could restrict the domain so that otherwise continuous functions are not continuous at their endpoints. 
A: You are correct that we can only define the limit of $\sqrt{x^2-16}$ as $x$ approaches $4$ from above, precisely because the expression is undefined for $x \lt 4$.
The proper notation for this in most books is $\lim_{x\to4^+} \sqrt{x^2-16}=0$.
In a basic college calculus course, the student will likely see a definition of limit very similar to the following, quoted from the Univ. of Tenn. Math Archives Visual Calculus by Larry Husch:

Definition.  The limit of $f(x)$ as $x$ approach $a$ is $L$
  $$ \lim_{x\to a} f(x) = L $$
  if and only if, given $\varepsilon \gt 0$, there exists $\delta \gt 0$ such that $0 \lt |x-a| \lt \delta$ implies that $|f(x)-L| \lt \varepsilon$.

This is somewhat casual in that no explicit requirement is made that $f(x)$ is defined anywhere.  My understanding would be that $f(x)$ must at least be defined on the deleted neighborhood of $a$ given by $0 \lt |x-a| \lt \delta$.  A Calculus I online tutorial does mention such a requirement:

Definition 1  Let $f(x)$ be a function defined on an interval that contains $x=a$, except possibly at $x=a$.  Then we say that,
  $$ \lim_{x\to a} f(x) = L $$
  if for every number $\varepsilon \gt 0$ there is some number $\delta \gt 0$ such that
  $$ |f(x)-L| \lt \varepsilon \;\; \text{ whenever } \;\; 0 \lt |x-a| \lt \delta $$

A pre-calculus text found through Google Books preview, Precalculus with Limits by Ron Larson, page 815, states the idea less formally:

Definition of Limit
  If $f(x)$ becomes arbitrarily close to a unique number $L$ as $x$ approaches $c$ from either side, then the limit of $f(x)$ as $x$ approaches $c$ is $L$.  This is written as $$ \lim_{x\to c} f(x) = L. $$ 

In my view even this least formal statement requires an approach to the limit from both sides of the point $x=c$. Of course English being as ambiguous as it is, someone wanting to resist this requirement might interpret matters as only requiring one side or the other, not necessarily both.  In mathematical practice, though, that weaker notion does not serve the purpose for which we want limits to be introduced, namely in order to define the derivative of a function at a point by taking the limit of the difference quotient.
