Proving $\lim_{x\to 1} x^3=1$ with $\epsilon$-$\delta$ definition

Question

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?

Answer 1

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...

Answer 2

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 $|x^2+x+1|<(1+\delta)^2+\delta+2$.

Answer 3

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, $$ \lim_{x\to 1}x^3=1. \;\blacksquare $$

Answer 4

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.

Answer 5

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 $\delta=\text{min}\{\frac{\epsilon}{7}, 1\}$. $\space\space\space\square$

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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.