Limit of $(x^3 - 3x^2 + 2x)$ as $x \to 1$ using epsilon delta definition

Question

I need to prove the following limit using epsilon delta method. I have come up this on my own just to practice skills.

$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$

So, I need to come up with some $\delta >0$ such that given $0 < \lvert x - 1 \rvert <\delta$, I need to prove that

$$ |(x^3 - 3x^2 + 2x) - 0| < \varepsilon \tag{1}$$

I can simplify eq $1$ as follows.

$$ |x| |x-1||x-2| < \varepsilon $$

When we have a non linear term, then we restrict $x$ to some distance from $1$. So, suppose we have $ 0 < |x-1| < 10$. This leads to $-10 < x-1 < 10$. Doing some algebra, we can get inequalities

$$ 0 \leqslant |x| < 11 \tag{2}$$

$$ 0 \leqslant |x-2| < 11 \tag{3}$$

From eq 2 and 3, it follows that

$$ 0 \leqslant |x||x-2| < 121 $$

And since $0 < |x-1| $, it follows that

$$ |x||x-1||x-2| < 121 |x-1| \tag{4}$$

To show that $|x||x-1||x-2| < \varepsilon$ , its sufficient to show that $ 121 |x-1| < \varepsilon$. And this will happen if $ |x-1| < \frac{\varepsilon}{121} $. So, this is another restriction on $|x-1|$. So, we can let

$$\delta = \text{min}\left(10, \frac{\varepsilon}{121}\right) $$

So, we can let this be our choice of $\delta$. Its clear that $\delta > 0$. So, now the official proof will follow.

Let $\varepsilon > 0$ be some arbitrary real. Suppose $0 < |x-1| < \delta$. With the choice of $\delta$ we have done, this means that $ 0 < |x-1| < 10$. As demonstrated above, equation 4 follows from this inequality

$$ |x||x-1||x-2| < 121 |x-1| $$

But now, we also have $ |x-1| < \frac{\varepsilon}{121} $. It means that $ 121 |x-1| < \varepsilon $. Using this, we get that

$$ |x||x-1||x-2| < \varepsilon $$

which can be rewritten as

$$ | (x^3 - 3x^2 + 2x) - 0 | < \varepsilon $$

Since, $\varepsilon > 0$ was arbitrary to begin with, its proven that

$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$

Is the proof correct ?

Answer 1

This proof is fine, but manipulating the bounded part can sometimes prove more challenging.

Here it is easy since it is only $|x||x-2|$, but often you will get benefits while shifting the limit towards $0$ by setting $x=x_0+u$ with $u\to 0$.

The main benefit is that limits in zero are part of your mathematical toolbox, also $|u|<\delta$ is easy to manipulate compared to $|x-x_0|<\delta$.

For instance here $|x(x-1)(x-2)-0|=\overbrace{|u|}^{<\epsilon}\overbrace{|1-u^2|}^{\le 1}<\epsilon$

Granted we take $\delta=\min(1,\epsilon)$ so that $\begin{cases}|u|<\delta<\epsilon\\0\le u^2<\delta^2<1^2\implies|1-u^2|\le 1\end{cases}$

The inequalities did appear more naturally than when you calculated the $10$ and $\frac 1{121}$ factors.

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Another advantage you will encounter in your epsilon-delta proofs is how it is easy to bound polynomials.

For instance imagine that in the problem instead of $|1-u^2|$ you had $|1-u^2+7u^3-2u^4|$.

Just proceed using the fact that for $|u|<1$ then $\cdots<|u|^4<|u|^3<|u|^2<|u|<1$.

The triangle inequality gives you $$|1-u^2+7u^3-2u^4|<|1|+|u^2|+7|u^3|+2|u^4|<1+1+7+2=11$$

Conclude by taking $\delta=\min(1,\frac{\epsilon}{11})$.

Answer 2

Other methode which easier

$f(x) =x^3 - 3x^2 +2x$

Suppose $|f(x)-l|<\epsilon $

Let's show $|x-1|<\alpha$

$|f(x)-l|=|f(x)-0|=|x^3 - 3x^2 +2x|<\epsilon $

$|x^3 - 3x^2 +2x|=|x^2-2x|×|x-1|<\epsilon $

Suppose $x\in [\frac{1}{2}, \frac{3} {2}]$

$\Rightarrow $$ \frac{1} {4}\leq x^2 \leq \frac{9}{4}$

And

$-3\leq-2x\leq - 1$

So :

$\frac{-11} {4}\leq x^2-2x \leq \frac{5}{4}<\frac{11}{4}$$\Rightarrow $$|x^2 - 2x|\leq\frac{11}{4}$

$\Rightarrow $$|x^2 - 2x||x-1|\leq\frac{11}{4}|x-1|$

We know

$|x^2-2x|×|x-1|<\epsilon$

So $\frac{11}{4}|x-1|<\epsilon $$\Rightarrow $$|x-1|<\frac{4\epsilon }{1 1} $

Pose

$\alpha =\frac{4\epsilon }{1 1} $

So we can see $|x-1|<\alpha$

Finally:

After the definition of limite we showed $lim_{x\to 1} x^3 - 3x^2 +2x=0$