Proving $\lim_{x\to 1} x^3=1$ with $\epsilon$-$\delta$ definition Problem: I need to formally prove that $$\lim_{x\to 1} x^3 = 1.$$
My work: This is what I have so far and I'm generally a bit stuck with these proofs from here onwards.
Because$$-\epsilon < x^3-1 < \epsilon  =  | x^3-1 | < \epsilon,$$
then
$$ -\epsilon < x^3-1 < \epsilon$$
$$-\epsilon+1 < x^3 < \epsilon +1$$
$$ \sqrt[3]{-\epsilon+1}< x < \sqrt[3]{\epsilon+1}.$$
Hoping that what I have done so far is correct. Am I right in thinking that
$$ \sqrt[3]{-\epsilon+1}< x < \sqrt[3]{\epsilon+1}$$
is giving me an interval where $x$ is going to give me a $f(x)$ value that falls within the distance $\epsilon$ from the limit on the $y$-axis ? Or is this interval smaller than the $\epsilon$-distance on the $y$-axis?
 A: Here is an easy way to prove it (let me know if a step doesn't make sense to you):
Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-1|<\delta$, then $|x^3-1|<\epsilon$. Now,
$$
|x^3-1| = |(x-1)(x^2+x+1)|.
$$
If $|x-1|<1$, that is, $-1<x-1<1$, then note that
$$
-1<x-1<1\Longleftrightarrow 0<x<2 \Longleftrightarrow x^2+x+1<2^2+2+1=7,
$$
and so 
$$
|x^3-1|=|x-1|(x^2+x+1)<7|x-1|.
$$
So if we take $\delta=\min\{1,\frac{\epsilon}{7}\}$, then
$$
0<|x-1|<\delta\Rightarrow |x^3-1|=|x-1|(x^2+x+1)<\frac{\epsilon}{7}\cdot 7 = \epsilon. 
$$
Thus, by the definition of a limit, 
$$
\lim_{x\to 1}x^3=1. \;\blacksquare
$$
A: After trying to understand the accepted answer, I have decided to rewrite the answer in my own words:
By the definition of limits, to show $\lim_{x \rightarrow 1}x^3=1$ we must show that for all $\epsilon > 0$ there exists $\delta > 0$ such that whenever $0 < |x-1|<\delta$ we have $|x^3-1| < \epsilon$.
Let $\epsilon > 0$, and let's look at the statement $|x^3-1| < \epsilon$. Notice that $|x^3-1|$ = $|x-1||x^2+x+1|$ and so the statement holds if $|x-1||x^2+x+1| < \epsilon$. At this point, we already have an $|x-1|$, and so we are close to choosing $\delta$ in terms of $\epsilon$, but we must get rid of $|x^2+x+1|$ first.
The statement holds if $|x-1|<\frac{\epsilon}{|x^2+x+1|}$. We would like to get only a constant on the right hand side, so is there anything we can say about $|x^2+x+1|$?
Notice that if we consider some number $r$, either $|x-1| \leq r$ or $|x-1| > r$. Let's arbitrarily choose $r=1$ (any choice should be fine here), and so either $|x-1| \leq 1$ or $|x-1| > 1$. We would need to show that our single choice of $\delta$ takes into account both possibilities.
Why is this useful? Let's look at the first possibility. We know that $|x-1| \leq 1$ holds if $-1 \leq x - 1 \leq 1$, which holds if $0 \leq x \leq 2$. But also, if $x$ lies in this range, we have $|x^2+x+1|\leq|2^2+2+1|=7$. So, $\delta=\frac{\epsilon}{7}$ would be a good choice so long as $|x-1| \leq 1$.
But, we need to find $\delta$ that works for the other values of $x$ also, since that was only one of the two cases for $x$!
However, we are free to choose the smallest $\delta$ we like. Can we make $\delta$ small enough where $|x-1|$ is never greater than $1$? That way we wouldn't even have to worry about the other case!
Notice that since $|x-1|<\delta$, to ensure $|x-1| \leq 1$ we just need to ensure $\delta < 1$.
Choose $\delta=\text{min}\{\frac{\epsilon}{7}, 1\}$. $\space\space\space\square$

This can probably become a common strategy: Make $\delta$ small enough so that the absolute value is less than an arbitrary real number $r$. Once that's the case, you can find an upper bound for $x$, and this allows one to get rid of all the extra $x$ terms.
A: You need to compute:
$$|x^3-1|=|(x-1)(1+x+x^2)|=|x-1|\times |1+x+x^2| $$
Now when $x$ is near $1$ (for instance $x\in [0,2]$) then $|1+x+x^2|\leq |1+2+2^2|=7$.
So you have :
$$|x^3-1|\leq 7|x-1|$$
Now given some $\epsilon$, I am sure you can guess $\delta$ such that...
A: Suppose $|x-1|<\delta$, 
then $|x^3-1|=|x-1||x^2+x+1|<\delta((1+\delta)^2+\delta+2)$
Now choose $\epsilon=\delta((1+\delta)^2+\delta+2)$.
Note: From $|x-1|<\delta$, we have $1-\delta<x<1+\delta$, then for $\delta \geq 1$,  $0\leq x^2<(1+\delta)^2$ and for $\delta \leq 1$, $(1-\delta)^2<x^2<(1+\delta)^2$
Thus itfollows that $|x^2+x+1|<(1+\delta)^2+\delta+2$. 
A: It is to show: $ \forall \epsilon \gt 0 \exists \delta \in R $ with:  $ |1-x^3| < \epsilon $ for $ |1-x| < \delta $
You get your $ \delta $ if you form this equation:$ |1-x^3| < \epsilon $ $ \ $  to:  $ \ $ $\sqrt[3]{1-\epsilon} < x < \sqrt[3]{1+\epsilon} $
Btw.: This line is wrong: $-\epsilon < x^3-1 < \epsilon  =  | x^3-1 | < \epsilon  $ 
$ \epsilon < \epsilon < \epsilon $ doesn't make too much sense. 
