Which is correct: negative infinity or 'does not exist'? For the $\lim_{x\to 10^-} ln(100-x^2)$, which is more correct?

$\lim_{x\to 10^-} ln(100-x^2)$ = negative infinity
$\lim_{x\to 10^-} ln(100-x^2)$ = DNE (Does not exist)

Graphically, $x$ approaches negative infinity, but by definition $ln(-n)$ is undefined. Does it matter whether I write the limit as negative infinity or simply non-existent?
 A: It's not that one is more correct than the other. One is right, the other is wrong. To see which one is correct, rewrite the limit like this:
$$\lim_{x \to 10^-}\ln(100 - x^2) = \lim_{x \to 10^-}\ln((10 - x)(10 + x))$$
You can see clearly that, as $x \to 10^-$
$$\begin{align}
10 - x &\longrightarrow 0^+\\
10 + x &\longrightarrow 20
\end{align}$$
Therefore $(10 - x)(10 + x) \longrightarrow 0^+$ and the limit is
$$\lim_{x \to 10^-}\ln(100 - x^2) = -\infty.$$
Of course $-\infty \notin \mathbb R$, but the limit notation is purely a shorthand for a longer notation.
A: It would be correct to say "does not exist in the real numbers".
It is common but incorrect to say "does not exist" when a concept is not defined.  Think about what "does not exist" actually means:
$$\lnot \exists z ~:~ z = \lim_{x \to 10^{-}} \ln(100 - x^2) \tag{A}$$
Can you prove that statement is true, using the definition of a limit?  If you use the definition of the limit in the form : "If $P(L, f, c)$, then $L = \lim_{x \to c} f(x)$" then (A) cannot be proven.  It is undefined.
If you use the definition of the limit of the form : "$P(L, f, c)$ if and only if $L = \lim_{x \to c} f(x)$" then you can prove that (A) is true.
A: There is more than one way a limit can fail to exist. One of the ways a limit can fail to exist is if it decreases without bound, or if it increases without bound, or on one side it increases without bound and on the other side it decreases without bound.
Here are other limits that fail to exist, for other reasons:
$$\lim_{x \to +\infty} \cos x$$
$$\lim_{x \to 0} \dfrac {|x|}x$$
$$\lim_{x \to 3} \begin{cases}1 & : x \text{ is rational} \\ 0 & : x \text{ is irrational} \\ \end{cases}$$
