$\sum_{n=1}^{\infty}a_{n}$ is formally the limit $\lim_{n\rightarrow\infty}s_{n}$
where $s_{n}=\sum_{k=1}^{n}a_{k}$.
In the case you mention ($x=0$) we have
$s_{n}=0$ for each $n$, hence $\lim_{n\rightarrow\infty}s_{n}=0$
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
Note that for any $x$ we have $x\cdot 0=0$ and therefore
$$\lim_{x\to\infty} (x\cdot 0) =\lim_{x\to\infty} 0=0$$