Q:

What makes a limit 'go away'?

According to this video  $$\lim\limits_{x\to\infty} \frac{11 - e^{-x}}{7} = \frac{11}{7}$$
I understand how this works, I don't understand the limit part though. I know $\lim\limits_{x\rightarrow \infty}{e^{-x}} = 0$, but what makes the limit disappear? I conclude that it disappears either because: 
A.
There are no more variables in the equation so no point to having a limit so we just start ignoring it
B.
Is has been used once so it goes away. 
Option B seems more likely, but it also it confusing because only one term seemed to be effected by the limit.

A:

The limit of a constant is just the value of constant, and when $\lim f(x)$ and $\lim g(x)$ both exist they satisfy
$$\lim(f(x)+g(x)) = \lim f(x) + \lim g(x)$$
$$\lim(f(x)g(x)) = \lim f(x) \cdot \lim g(x)$$
In other words, here, you have
$$\lim \frac{11-e^{-x}}{7} = \lim \frac{1}{7} \cdot \lim(11-e^x) = \lim \frac{1}{7}(\lim 11 - \lim e^{-x})$$
Since $\frac{1}{7}$ and $11$ are constant, $\lim e^{-x} = 0$, you get
$$\lim \frac{11-e^{-x}}{7} = \frac{1}{7} \cdot (11 + 0) = \frac{11}{7}$$
(I've left the subscript $x \to \infty$ off the limit signs.)

A:

Option (A) is almost right!
Specifically, it is a property of limits that if $c$ is a constant,
\begin{equation*}
\lim_{x\to a}{c}=c.
\end{equation*}
You can prove this using the definition of a limit (that is, using a $\delta$-$\epsilon$ proof).
So $\lim_{x\to\infty}{\frac{11}{7}}=\frac{11}{7}$, having no limit on the right-hand side, just like $\lim_{x\to\infty}{e^{-x}}=0$, having no limit on the right-hand side.

A:

If we continuously extend the real exponential to have values $e^{+\infty} = +\infty$ and $e^{-\infty} = 0$, then this can be seen as taking the limit of a continuous function, which can be done by simply plugging in the limiting value:
$$\lim_{x\to +\infty} \frac{11 - e^{-x}}{7} = \frac{11 - e^{-(+\infty)}}{7} 
= \frac{11 - e^{-\infty}}{7} = \frac{11 - 0}{7} = \frac{11}{7} $$

