Is my proof of $\lim _{ x\rightarrow \infty }{ \frac { x+8 }{ x+3 } =1 } $ correct? $$\lim _{ x\rightarrow \infty  }{ \frac { x+8 }{ x+3 } =1 } $$
Proof: Let $\epsilon > 0$
$$\left| \frac { x+8 }{ x+3 } -1 \right| <\epsilon $$
$$\Longrightarrow \left| \frac { x+8 }{ x+3 } -\frac { x+3 }{ x+3 }  \right| <\epsilon $$
$$\Longrightarrow \left| \frac { 5 }{ x+3 }  \right| <\epsilon $$
$$\Longrightarrow \frac { 5 }{ \left| x+3 \right|  } <\epsilon $$
Without loss of generality, assume that $x>-3$
Then, we get: $$\frac { 5 }{ x+3 } <\epsilon $$
$$\Longrightarrow 5< \epsilon (x+3)$$
$$\Longrightarrow 5<\epsilon x+3\epsilon $$
$$\Longrightarrow \epsilon x>5-3\epsilon $$
$$\Longrightarrow x>\frac { 5 }{ \epsilon  } -3$$
This concludes my proof, but I still have so many questions. I feel like a computer when I'm doing this. I don't truly understand what I'm doing and I know that this will come back to haunt me at some point. 
So, some of the things that I want clarification on:
1) If my proof is even correct and complete
2) Why I should assume that $x>-3$ It seems very arbitrary.
3) What exactly this all means and why did it prove what I was asked.
 A: Your proof has the right idea, but proceeds in the wrong direction. In particular, to show that $\lim_{x\rightarrow\infty}f(x)=c$ you need that there is some $N$ so that for any $x>N$ we have $|f(x)-c|<\varepsilon$. Now, you've got all the elements together, but what you have is:

If $|f(x)-c|<\varepsilon$ then $x>N$.

which is distinct from what you need:

If $x>N$ then $|f(x)-c|<\varepsilon.$

Now, all of your steps are reversible, so you can just write your proof in reverse and be correct. Notice that your proof implicitly sets $N=\frac{5}{\varepsilon}-3$, which is why that numbers is important.
Also, as to why you assume $x>-3$ that's just so that you don't have to deal with the vertical asymptote at $x=3$. You can do that since you only care about how "large" $x$ act.
A: To assert 
$$\lim _{ x\rightarrow \infty  }{ \frac { x+8 }{ x+3 } =1 }$$
is to say, in casual words, that if $x$ is large, that the fraction is close to $1$. In math legalese, it is to say that given an $\epsilon >0$, one can find an $L$ such that $x > L$ implies
 $$ \left| \frac { x+8 }{ x+3 } -1 \right| <\epsilon.$$
Your algebra is really calculating $L$:
$$ L = {\frac { 5 }{ \epsilon  }} -3.$$
So you really want to convince the reader, that if he or she starts with the last line of your argument, and read it backwards, (i.e., with your arrow $\Longrightarrow$ replaced with $\Longleftarrow$, so that a given line implies the preceding), they arrive at the conclusion that the fraction is within $\epsilon$ of $1$.
On the other hand, you figured out your $L$ in the order that you wrote down your argument - of course... So either you use $\Longleftrightarrow$ (so both you and your reader are happy), or you secretly do, on the side, your calculation, and write the argument, starting with if
$  x> {\frac { 5 }{ \epsilon  }} -3$, then ...., concluding with the "we are forced to conclude the fraction is within  $\epsilon$ of $1$ - proving that the limit is $1$, as desired." 
OK?  
