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: 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:
$$\lim_{x\to +\infty} \frac{11 - e^{-x}}{7} = \frac{11 - e^{-(+\infty)}}{7} 
= \frac{11 - e^{-\infty}}{7} = \frac{11 - 0}{7} = \frac{11}{7} $$
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: I can guess that your doubt comes from a first step that you do in your mind that goes like this: "$\lim_{x\rightarrow \infty}{e^{-x}} = 0$, so I can replace $e^{-x}$ with $0$ in the text and I remain with
$$
\lim_{x\to ∞} \frac{11-0}{7},
$$
then what happens to the limit in this last equation?"
This reflects a common misunderstanding. Repeat after me: in a limit, one cannot replace arbitrary subexpressions with their limit. There is no theorem that says that. If you've been doing that, you are wrong; change your habits.
For an easier example when this doesn't work, $\lim_{x\to 0} x = 0$, but
$$
\lim_{x\to 0} \frac{x+x^3}{x^2} \neq \lim_{x\to 0} \frac{0+x^3}{x^2}.
$$
Clive Newstead's and Ilmari Karonen's answers show some alternative manipulations that are legitimate by theorems.
A: As EPAstor notes, your reason (A) is closer to the truth, but it's not the complete answer.
The reason why $\displaystyle \lim_{x \to \infty} e^{-x} = 0$ implies $\displaystyle \lim_{x \to \infty} \frac{11 - e^{-x}}{7} = \frac{11}{7}$ is that
$$\lim_{x \to \infty} \frac{11 - e^{-x}}{7}
= \frac{\lim_{x \to \infty} 11 - e^{-x}}{7}
= \frac{11 - (\lim_{x \to \infty} e^{-x})}{7}
= \frac{11 - 0}{7}
=  \frac{11}{7}$$
and the reason we can move the limit inside the fraction like that is that the functions
$$z \mapsto 11 - z \quad \text{and} \quad z \mapsto \frac z 7$$
are both everywhere continuous.  It is a rather straightforward consequence of the definition of continuity (prove it!) that, if a function $f$ is continuous at $c = \displaystyle\lim_{x \to a} z(x)$, then
$$\lim_{x \to a} f(z(x)) = f(\lim_{x \to a} z(x)) = f(c).$$
A: First note that
$$ \lim\limits_{x\to\infty} f(x)= f\left(\lim\limits_{x\to\infty} x\right) $$
And 
$$ \lim\limits_{x\to\infty} e^{-x}= \lim\limits_{x\to\infty} \frac{1}{e^{x}} =0$$
Therefore 
$$\lim\limits_{x\to\infty} \frac{11 - e^{-x}}{7} = \frac{11 - \lim\limits_{x\to\infty} e^{-x}}{7} =\frac{11}{7}$$
