Let $f$ be the characteristic function of the set $$A:=\left\{\left({1\over n},e^{-n}\right)\>\Biggm|\>n\geq 1\right\}\subset{\mathbb R}^2\ .$$
Lemma. Any curve
$$t\mapsto\cases{x(t):=t\>p(t)&\cr y(t):=t\>q(t)&\cr }\qquad(t\geq 0)\tag{1}$$
with $p$ and $q$ real analytic functions, $p(0)$ and $q(0)$ not both zero, can contain only finitely many points of $A$.
Proof. Note that the points of $A$ are lying on the curve
$$\epsilon:\quad y=e^{-1/x}\qquad(x>0)\ .$$
Let $\gamma$ be any curve $(1)$. If $p(0)=0$ then $|y(t)|>|x(t)|$ for all sufficiently small $t$, while $|y|<|x|$ on $\epsilon$. When $p(0)\ne0$ we can eliminate $t$ from $(1)$ and write $\gamma$ as graph of a real analytic function: $y(x)\equiv0$ or
$$ y= x^r(a_0+a_1x+a_2x^2+\ldots)\qquad(0\leq x<\rho)\ ,$$
where $r\geq1$ and $a_0\ne0$. In the second and more dangerous case there is an $h>0$ with
$$|y(x)|>{|a_0|\over 2}x^r>e^{-1/x}\qquad(0<x<h)\ .$$
Therefore the curve $\gamma$ cannot contain any points of $A$ with ${1\over n}<h$.
From this Lemma it follows that
$$\lim_{t\to 0+}f\bigl(x(t),y(t)\bigr)=0$$
for all such curves, even though $f$ is not continuous at $(0,0)$.
So you've stumbled upon the concept of a limit and how it can be different than evaluating an expression at the value you are approaching.
When we are asking, for example, what the limit of $2 \cdot\frac{x-1}{x-1}$ is as $x$ approaches $1$, what we aren't asking for is what the value of the expression $2 \cdot\frac{x-1}{x-1}$ is at $x = 1$. This expression is undefined at $x = 1$ since you get $\frac{0}{0}$.
But what we want to know is: as you choose $x$ closer and closer to the value $1$, are the values of $2 \cdot\frac{x-1}{x-1}$ getting closer and closer to some value?
Well, as it turns out in my example, it's easy to see that, yes, the values are getting closer to some value. If $x$ is not equal to $1$, $2 \cdot\frac{x-1}{x-1}$ is equal to $2$. So as $x$ gets closer and closer $1$, the expression $\frac{x-1}{x-1}$ is "getting closer and closer" to $2$, since it's already always $2$ for all $x$, which implies it's always $2$ for all $x$ near $1$ (except of course at $x=1$).
So a limit doesn't care about what happens at the value (e.g., at $x=1$). It cares about what the expression looks like as $x$ gets closer and closer to $1$.
Sometimes, evaluating an expression agrees with its limit. For example, consider the expression $x/2$. What happens as $x$ gets closer and closer to $0$? Well, if $x$ is getting really small, so is $x/2$. So when $x$ is near $0$, $x/2$ is near $0$. So we say $\lim \limits_{x \to 0} x/2 = 0$. But if you evaluate $x/2$ at $x=0$, then we also get $0$. When this happens, we say the expression is continuous at $x = 0$. So $f(x) = x/2$ is continuous at $x = 0$.
In the case of your example, $(k/h) = 128 + 16h$ when $h$ is not zero. So, since the $(k/h)$ equals $128 + 16h$ for all non-zero $h$, then as $h$ gets smaller, $(k/h)$ behaves as $128 + 16h$. But $128 + 16h$ is getting closer and closer to $128$ as $h$ goes to $0$ since $16h$ is getting smaller and smaller. That means $k/h$ is getting closer and closer to $128$. So we write $\lim \limits_{h \to 0} k/h = 128$. Note that this doesn't mean $k/0 = 128$. As I discussed above, we don't care about what happens at the value of $h = 0$. We only care about what happens as $h$ gets closer and closer to $0$. $k/0$ is undefined. But $\lim \limits_{h \to 0} k/h = 128$.
Best Answer
First of all, if absolute $0$ is the absolute value of $0$, then note that $\lvert0\rvert=0$. So, there is no difference between zero and absolute zero.
On the other and, if you write $x^2$ as $e^{2\ln x}$, then you have a problem: since$\ln x$ doesn't exist when $x<0$, from the fact that $\lim_{x\to0}e^{2\ln x}=0$ all you can deduce is that $\lim_{x\to0^+}x^2=0$. Of course, since $x^2$ is an even function, it follows from this that $\lim_{x\to0^-}x^2=0$ too.