Limiting formula for differentiable function

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem and solution,

I'm confused how $x \in (c - \delta, c + \delta)$ is the same as $0 <| x - c| <\delta$.


I think it is the same as $c - \delta < x < c + \delta$ which we break into parts $c - \delta < x \implies \delta > -(x - c)$ and $x < c + \delta \implies x - c < \delta$. Thus recombining the two inequalities using the definition of absolute value we get $| x - c| < \delta$ don't we please?

Thanks for any help!

 

[anuttarasammyak]

Homework Statement: Please see below
Relevant Equations: Please see below

I'm confused how x∈(c−δ,c+δ) is the same as 0<|x−c|<δ.

$$c-\delta<x<c+\delta$$
$$-\delta<x-c<\delta$$

 

[member 731016]

$$c-\delta<x<c+\delta$$
$$-\delta<x-c<\delta$$View attachment 347061

Thank you for your reply @anuttarasammyak !

But ain't $$c-\delta<x<c+\delta$$ same as $| x - c| <\delta$?

Thanks!

 

[anuttarasammyak]

But ain't c−δ<x<c+δ same as |x−c|<δ?

Do you observe that from c−δ<x<c+δ,
$$-\delta<x-c<\delta$$
by adding -c to all the terms ?

 

[member 731016]

Do you observe that from c−δ<x<c+δ,
$$-\delta<x-c<\delta$$
by adding -c to all the terms ?

Thank you, yes!

 

[pasmith]

$|x - c| < \delta$ means that $x$ is at most $\delta$ away from $c$. Thus, $c - \delta < x < c + \delta$. Of course it also means that $c$ is at most $\delta$ away from $x$, so that $x - \delta< c < x + \delta$.

 

[WWGD]

Notice , for $x \neq 0$, $|x|>0$. So you're only excluding the option $x=c$.