Simple epsilon-delta question

Question

Is it true that for every $ε > 0$, there is $δ > 0$, such that $0 < |x−2| < δ ⇒ |(x^2 −x)−2| < ε$?

Now I know that $|(x^2 −x)−2|$ is same as $|(x-2)(x+1)|$, but I am not sure how to link that with the first bit of info given. In general epsilon-delta proofs confuse me.

So I start by saying that there is an epsilon s.t $|(x^2 −x)−2| < ε$. And if this is true then there is a delta s.t $0 < |x−2| < δ$. Or is it the other way around?

Now, if $|(x^2 −x)−2| < ε$ then $|(x-2)(x+1)| < ε$ and $|x-2||x+1| < ε$ and $$|x-2|<\frac{ε}{|x+1|}$$ But since epsilon is always positive and so is $|x+1|$ then a delta always exists.

Is my proof correct or totally wrong? I feel as though all I have done is rearranged the equation, and not really proved anything.

It should be $x^2$, I guess. @Anna Please edit by writing x^2 — Friedrich Philipp, Mar 13 '16 at 05:46
This question must be edited and rewritten more clearly, otherwise assistance is very difficult. — Sarvesh Ravichandran Iyer, Mar 13 '16 at 05:49
@ChristopherHalverson Why doesn't it? It is about showing that $x^2-x$ is continuous at $x=2$. — Friedrich Philipp, Mar 13 '16 at 05:51

score 1 · Answer 1 · answered Mar 14 '16 at 14:21

Let the $\epsilon = \epsilon_0$ satisfying $|x^2-x-2| < \epsilon_0$. Initially choose $\delta$ to be $1$. We will refine this delta.

$-\epsilon_0 < x^2-x-2 < \epsilon_0$

$\implies -\epsilon_0+\frac{9}{4} < x^2-x+\frac{1}{4} < \epsilon_0+\frac{9}{4}$

$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 < \epsilon_0+\frac{9}{4}$

$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 $

Now, if $\epsilon_0 < \frac{9}{4}$

$\implies \sqrt{-\epsilon_0+\frac{9}{4}} < x-\frac{1}{2} $

$\implies \sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2} < x+1 $

Remember we had $|x^2-x-2| < \epsilon_0$. Then

$-\epsilon_0 < (x-2)(x+1) < \epsilon_0$

$\implies -\frac{\epsilon_0}{x+1} < (x-2) < \frac{\epsilon_0}{x+1}$

$\implies |x-2| < \frac{\epsilon_0}{x+1} < \frac{\epsilon_0}{\sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2}}$.

Therefore choose $\delta = min( 1 , \frac{\epsilon_0}{\sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2}} )$.

(note that for $\epsilon_0 > \frac{9}{4}$, choose $\delta$ as if $\epsilon_0 = \frac{9}{4}$, this will eventually satisfy the condition)

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

I always like to refer people to my answer here when it comes to simple polymonial $\delta$-$\epsilon$ proofs. Read this link so that you understand my methodology here.

Scratch work:

$$|x^2-x-2| = |(x-2)(x+1)| = |x-2||x+1|\text{.} $$ Take $\delta = 1$. Then $$|x-2| < 1 \Longleftrightarrow -1 < x-2 < 1 \Longleftrightarrow 1 < x < 3 \Longleftrightarrow 2 < x+1 < 4 \implies |x+1| < 4\text{.}$$ So for $\epsilon > 0$, $$|x-2||x+1| < 4|x-2| < 4\delta = 4\left(\dfrac{\epsilon}{4}\right)=\epsilon$$ if $\delta = \dfrac{\epsilon}{4}$, so we choose $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$.

Proof:

Let $\epsilon > 0$ be given. Choose $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$. Then $$|(x^2-x)-2| = |x^2-x-2| = |x-2||x+1| < 4|x-2|$$ (since if $|x-2| < 1$, $|x+1| < 4$), and $$4|x-2| < 4\delta \leq 4\left(\dfrac{\epsilon}{4}\right) = \epsilon\text{.}$$

score 0 · Answer 3 · answered Mar 14 '16 at 15:00

Here are some general results which enable us to handle the Q of continuity for a broad class of real functions:

(1)... Constant functions are continuous.

(2)... f(x)=x is continuous

(3)...f(x)=|x| is continuous.

(4)...For continuous f, g :

... (i)... h(x)=f(x)+g(x) is continuous.

... (ii)...j(x)=f(x)g(x) is continuous.

... (iii)...k(x)=f(g(x)) is continuous.

These are readily proven by the standard $\epsilon , \delta$ method. One immediate consequence is that a polynomial $p(x)$ and its absolute value $|p(x)|$ are continuous functions

Simple epsilon-delta question

3 Answers3