Question about minimum function and limit definition (proper)

[Amad27]

Hi,

Suppose you want to prove $|x - a||x + a| < \epsilon$

You know

$|x - a| < (2|a| + 1)$
You need to prove

$|x + a| < \frac{\epsilon}{2|a| + 1}$

So that

$|x - a||x + a| < \epsilon$

Why does Michael Spivak do this:

He says you have to prove --> $|x + a| < min(1, \frac{\epsilon}{2|a| + 1})$ in order to finally prove, $|x + a||x - a| < \epsilon$

Why do we need the $min$ function there?

Thanks!

 

[mathbalarka]

OK, let's write down the statement of the problem

Claim $|x+a| < \text{min}(1, \epsilon \cdot (2|a|+1)^{-1}) \Rightarrow |x+a||x-a| < \epsilon$

If $\text{min}(1, \epsilon \cdot (2|a|+1)^{-1}) = 1$, $\epsilon > 2|a|+1$, and we know $2|a|+1 > |x-a|$, implying $|x-a| < \epsilon$. As $|x+a| < \text{min}(1, \epsilon(2|a|+1)^{-1}) = 1$, $|x-a||x+a| < \epsilon \cdot 1 = \epsilon$ $\blacksquare$

I'm not sure if you really want to "prove" that $|x+a||x-a| < \epsilon$, or find a suitable $\epsilon$ to do so, or whatever you have in mind so I don't think I can help you more that this if you can't give the context. What was the original problem?

 

[Amad27]

OK, let's write down the statement of the problem

Claim $|x+a| < \text{min}(1, \epsilon \cdot (2|a|+1)^{-1}) \Rightarrow |x+a||x-a| < \epsilon$

If $\text{min}(1, \epsilon \cdot (2|a|+1)^{-1}) = 1$, $\epsilon > 2|a|+1$, and we know $2|a|+1 > |x-a|$, implying $|x-a| < \epsilon$. As $|x+a| < \text{min}(1, \epsilon(2|a|+1)^{-1}) = 1$, $|x-a||x+a| < \epsilon \cdot 1 = \epsilon$ $\blacksquare$

I'm not sure if you really want to "prove" that $|x+a||x-a| < \epsilon$, or find a suitable $\epsilon$ to do so, or whatever you have in mind so I don't think I can help you more that this if you can't give the context. What was the original problem?

Hi,

Thanks for replying, this could help very much.

The original problem is proving.

as $x -> a, f(x) = x^2 -->a^2$

We can also use $\delta$ if that would help by any chance.

How do you know:

$$2|a| + 1 > |x - a|$$?

He then uses we need to prove

$$|x - a||x + a| < \epsilon$$
To do that he says,

$$\text{if}\space |x + a| < \text{min}(1, \frac{\epsilon}{2|a| + 1})$$

What is the point of $\text{min}$ what does that even imply?

Why do we require the $1$ there? Why not directly prove

$$|x + a| < \frac{\epsilon}{2|a| + 1}$$

Thanks!

 

[mathbalarka]

By definition of continuity, you have to prove that for every $\varepsilon > 0$, there is always a $\delta > 0$ such that if $0 < |x - a| < \delta$ then $|x^2 - a^2| < \epsilon$.

Assume $\delta < 1$. Then $|x - a| < 1 \Longrightarrow -1 < x - a < 1 \Longrightarrow -1 + a < x < 1 + a$ which implies $|x + a| < 2|a| + 1$. Hence $|x^2 - a^2| = |x + a||x - a| < (2|a| + 1)|x - a| < \epsilon$. Then $|x - a| < \epsilon \cdot (2|a|+1)^{-1}$.

Can you see why $\delta$ is picked to be the minimum of $1$ and $\epsilon \cdot (2|a|+1)^{-1}$? Verify if nessesary.

 

[Amad27]

By definition of continuity, you have to prove that for every $\varepsilon > 0$, there is always a $\delta > 0$ such that if $0 < |x - a| < \delta$ then $|x^2 - a^2| < \epsilon$.

Assume $\delta < 1$. Then $|x - a| < 1 \Longrightarrow -1 < x - a < 1 \Longrightarrow -1 + a < x < 1 + a$ which implies $|x + a| < 2|a| + 1$. Hence $|x^2 - a^2| = |x + a||x - a| < (2|a| + 1)|x - a| < \epsilon$. Then $|x - a| < \epsilon \cdot (2|a|+1)^{-1}$.

Can you see why $\delta$ is picked to be the minimum of $1$ and $\epsilon \cdot (2|a|+1)^{-1}$? Verify if nessesary.

I see an issue though.

What about for all those that

$$|x - a| > 1$$

Why not consider those values as well?

 

[Amad27]

I see an issue though.

What about for all those that

$$|x - a| > 1$$

Why not consider those values as well?

Or wait,

Why do you need to bound $|x - a|$ anyway?

AND

What if $\epsilon/2|a| + 1 < 1$

Then what happens?