How can I use a precise definition to find values of delta that correspond with given epsilon values I have been given this problem:
For the limit $$\lim_{x\to 2}({x^3-3x+4})=6$$
illustrate "Definition 2" (I have included this below) by finding values of $\delta$ that correspond to $\varepsilon=0.2$ and $\varepsilon=0.1$
"Definition 2:" My textbook says to this definition "Let $f$ be a function defined on some open interval that contains the number $a$, except possibly $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$\lim_{x\to a}f(x)=L$$
if for every number $\varepsilon>0$ such that $$if\;0<|x-a|<\delta\;\;\;then\;|f(x)-L|<\varepsilon$$
This has been a seriously frustrating problem, I've been working on it for two days, I just don't understand this concept at all. The only thing I think I understand is that the values of $\delta$ correspond to values on the x-axis, and $\varepsilon$ is related to the limit. 
How do I go about solving this problem? 
 A: I think you're close to getting it to "click". What you said at the end is correct: the $\delta$ value represents the distance between $x$ and $a$, and the $\epsilon$ value represents the distance between $f(x)$ and the limit $L$.
So the $\delta$, $\epsilon$ limit definition is a quantification of the statement "As $x$ becomes closer to $a$, $f(x)$ becomes closer to $L$".
For your problem at hand, you are given $\epsilon = 0.1$. So what they're asking is "How close does $x$ have to be to $a = 2$ before $f(x) = x^3 - 3x + 4$ is at most $0.1$ away from $L = 6$?"
A: These kinds of problems are often done backwards. We start with what we want.
$$
|x^3-3x+4 -6|<\varepsilon
$$
for $x$ near $2$. So replace $x$ with $2+\delta$ for some $|\delta|$ small.
Now, we can start simplifying the left. 
$$
|(2+\delta)^3-3(2+\delta)-2|=|8+4\delta+2\delta^2+\delta^3-6+\delta-2|=|\delta^3+2\delta^2+5\delta|
$$
So what $\delta$ makes $|\delta^3+2\delta^2+5\delta|<\varepsilon$? Factoring $\delta$ gives
$$
|\delta|\;|\delta^2+2\delta+5|
$$
If $\delta$ is already less than 1/3, then the right factor is less than 6. So we have we want to choose $\delta$ so that $\delta 6<\varepsilon$
