Limits and common sense I'm stuck in understanding of limits.
It all makes sense, but at a certain point my answers which seem logical to me are not true.
Please can somebody explain why as a huge number gets divided by a big number plus the relatively small value $1$, the result gets closer and closer to zero:
$$
lim_{x\to \infty} \frac{\infty!}{(\infty + 1)!} =0.
$$
The next one is not equal zero as well.
$$ 
lim_{x\to \infty} \frac{\infty}{\infty-8}=1
$$
The relatively huge number gets divided by another relatively huge number minus 8, which is so little we could ignore it. Why is it approaching 1 then? 
It can make sense from, I don't know, a visual point of view where 
$$ 
lim_{x\to \infty} \frac{\infty}{\infty-8}=\frac{\infty}{\infty}=1
$$
but I could as well do that with first one and also get 1, which is wrong.
 A: This answer is divided in two parts. In the first part I will share some of the theory behind infinity, and in part $2$ we will look at how to use this theory to solve limits.
Part 1 (Theory)
Before we start, I wanna make a comment on something you said. You said in a comment that you think we deal with same-size infinities all around. This is not true. Consider two limits
$$\lim _{x \to \infty} x = \infty$$
$$\lim _{x \to \infty} 2x = \infty$$
While we use the same symbol for the results of both of these limits, the infinity resulting from $2x$ is in some sense "bigger" or  (as I like to think about it) "stronger" than the infinity resulting from $x$. I think this is the root cause of your confusion, and it's the reason why in your question, one limit is equal to $0$, while another is equal to $1$. Anyway, here's the intro to the theory:

The symbol $\infty$ is not a number. Technically it's an extended real number. But no need to get into technicalities. Going by intuition is best.
The symbol $\infty$ behaves like a number in some cases, but not in others.
Here are some of the things you can always do with infinity. These should all be very clear intuitively. If they are not, let me know:
A) $\infty + a = a + \infty = \infty$ for any number $a$
B) $\infty + \infty = \infty$
C) $\frac {a}{\infty} = 0$ for any number $a$
D) $\frac {\infty}{a} = \pm \infty$, depending on the sign of $a$.
E) $\infty \cdot \infty = \infty$
F) $\infty \cdot a = a \cdot \infty = \pm \infty$, depending on the sign of $a$.
G) $\infty^2 = \infty, \infty^3 = \infty,$ $\infty^4 = \infty$, etc.
If you understand these intuitively, there should be no doubt that the above are true. I did not include everything here, but if you understand these intuitively, you should hopefully be able to come up with more of these by yourself (another one: $2^{\infty} = \infty$).
The following operations with $\infty$ are undefined (in the same way that, for example, $\frac 10$ is undefined).

*

*$\infty - \infty$

*$\frac {\infty}{\infty}$
Why is this? Consider for example $\infty + \infty$ compared to $\infty - \infty$. In the first expression, there is no doubt that the result should be $\infty$. If you add two huge numbers, you get a huge number. The huge numbers
are working together to make an even bigger number.
But for the second expression, there is doubt. If you subtract two really huge numbers, it totally depends on how big the two numbers are compared to each other. For example, $100000 - 1000$ is really big, but $1000 - 1000000$ is really small, and $100000 - 100000$ is zero. The two huge numbers are working against each other. The first number wants to pull the result up, but the second number wants to pull the result down. It all depends on which huge number is "stronger".
Here are examples with limits which demonstrate that with $\infty - \infty$, anything can happen, depending on how "strong" the specific infinities involved are. In their original form, all of these limits are of the type $\infty - \infty$ which is undefined. But after some easy manipulations, that are not of that type anymore, and we can compute them easily.

*

*$\displaystyle\lim_{x \to \infty} (5x - x) = \lim_{x \to \infty} 4x = 4 \times \infty = \infty$

*$\displaystyle\lim_{x \to \infty} [(x+4) - x] = \lim_{x \to \infty} 4 = 4$

*$\displaystyle\lim_{x \to \infty} (x - x) = \lim_{x \to \infty} 0 = 0$

*$\displaystyle\lim_{x \to \infty} [(x-4) - x] = \lim_{x \to \infty} -4 = -4$

*$\displaystyle\lim_{x \to \infty} (x - 5x) = \lim_{x \to \infty} -4x = -4 \cdot \infty = - \infty$
These may seem like silly examples, but they illustrate the point.
Same story with $\frac {\infty}{\infty}.$ The huge number in the numerator wants to pull the result up (because for example $\frac {\infty}{5} = \infty$) but the huge number in the denominator wants to pull the result down towards zero (because for example $\frac {5}{\infty} = \infty$). Again, the result will depend on which huge number is "stronger".
Here are examples which illustrate that again, anything can happen, depending on the "strength" of the specific infinities involved.

*

*$\displaystyle \lim_{x \to \infty} \frac {x^2}{x} = \lim_{x \to \infty} x = \infty$


*$\displaystyle \lim_{x \to \infty} \frac {4x}{x} = \lim_{x \to \infty} 4 = 4$


*$\displaystyle \lim_{x \to \infty} \frac {x}{x^2} = \lim_{x \to \infty} \frac 1x =  \frac {1}{ \infty} = 0.$
$\\$
Part 2 (Application)
So, how do you actually apply this stuff to solve a limit of the form $\lim_{x \to \infty} f(x)$? You start by plugging in $\infty$, and see what happens. There are two possibilities:
1) Once you plug in, you only have allowable operations to do. In this case, you do those operations, and get the answer. For example
\begin{align*}
 \lim_{x \to \infty} \dfrac {2x^3 + 3x - 4}{5} = \dfrac {2 \cdot \infty^3 + 3\cdot \infty - 4}{5} = \dfrac {\infty + \infty - 4}{5} = \frac{\infty}{5} = \infty
\end{align*}
2) Once you plug in, you get one of the forbidden operations, such as $\frac {\infty}{\infty}.$ If this happens, then you need to go back to the original and try to "fix" the issue. Either you do an algebraic manipulation, or (more advanced) use L'Hopital's Rule, or (even more advanced) Taylor series, etc.
Once you have done the manipulation (or whatever else) then you plug in $\infty$ again, and hope that you end up in case $1$. (notice that this is exactly what I did in the examples above when showing that $\infty - \infty$ or $\frac {\infty}{\infty}$ could be anything).If you end up in case $2$ again, then you repeat.
An example:
$$\lim_{x \to \infty} \frac {x^2 - 4}{x-2}$$
Plugging in gives us $\frac {\infty^2 - 4}{ \infty - 2} = \frac {\infty}{\infty }$ which is undefined, so we go back to the original and manipulate.
$$\lim_{x \to \infty} \frac {x^2 - 4}{x-2} = \lim_{x \to \infty} \frac {(x-2)(x+2)}{x-2} = \lim_{x \to \infty} x + 2 = \infty + 2 = \infty.$$

Now, this is what happens in your two examples in the question. Presumably, the first question started as
$$\lim_{n \to \infty} \dfrac {n!}{(n+1)!}$$
here we think of $n$ as a whole number, because factorials are only defined for whole numbers.
If you plug in $\infty$, you end up with $\frac{\infty}{\infty}$. So we go back and manipulate:
$$\lim_{n \to \infty} \dfrac {n!}{(n+1)!} = \lim_{n \to \infty} \dfrac {1 \cdot 2 \cdot 3 \cdots \cdot n}{1 \cdot 2 \cdot 3 \cdots \cdot n \cdot (n+1)!} = \lim_{n \to \infty} \frac {1}{n+1} = \frac {1}{\infty + 1} = \frac {1}{\infty} = 0.$$
In your second question. Presumably the problem started as
$$\lim_{x \to \infty} \frac {x}{x-8}$$
If you plug in you get $\frac {\infty}{\infty}$. So we go back and manipulate.
$$\lim_{x \to \infty} \frac {x}{x-8} = \lim_{x \to \infty} \frac {x}{x(1-\frac 8x)} = \lim_{x \to \infty} \frac {1}{1-\frac 8x} = \frac {1}{1 - \frac {8}{\infty}} = \frac {1}{1-0} = 1$$
These manipulations (especially for the second problem) might seem like they come out of nowhere. But if you study limits, you will see the same 3-4 types of manipulations over and over, and you will know how to use them. The ones I used are very common.
A: Well when you have limits as $x$ approaches infinity of a quotient. The limit of the quotient is the limit of the highest power in the numerator over the highest power in the numerator. So 
for example in your second case:
$\lim_{x\rightarrow\infty}\frac{x}{x-8}=\lim_{x\rightarrow\infty}\frac{x}{x}=1$
For your first case: $\lim_{x\rightarrow\infty}\frac{x!}{(x+1)!}=\lim_{x\rightarrow\infty}\frac{x!}{(x+1)x!}=\lim_{x\rightarrow\infty}\frac{1}{x+1}=0$
Edit: You shouldn't be treating infinities as numbers because there are different kinds of infinities.
