Limiting formula for differentiable function

member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this problem and solution,
1718654014506.png


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!
 
Physics news on Phys.org
ChiralSuperfields said:
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&lt;x&lt;c+\delta
-\delta&lt;x-c&lt;\delta
1718684755906.png
 
Last edited:
  • Love
Likes member 731016
anuttarasammyak said:
c-\delta&lt;x&lt;c+\delta
-\delta&lt;x-c&lt;\deltaView attachment 347061
Thank you for your reply @anuttarasammyak !

But ain't c-\delta&lt;x&lt;c+\delta same as ##| x - c| <\delta##?

Thanks!
 
ChiralSuperfields said:
But ain't c−δ<x<c+δ same as |x−c|<δ?
Do you observe that from c−δ<x<c+δ,
-\delta&lt;x-c&lt;\delta
by adding -c to all the terms ?
 
anuttarasammyak said:
Do you observe that from c−δ<x<c+δ,
-\delta&lt;x-c&lt;\delta
by adding -c to all the terms ?
Thank you, yes!
 
|x - c| &lt; \delta means that x is at most \delta away from c. Thus, c - \delta &lt; x &lt; c + \delta. Of course it also means that c is at most \delta away from x, so that x - \delta&lt; c &lt; x + \delta.
 
  • Love
Likes member 731016
Notice , for ##x \neq 0##, ##|x|>0##. So you're only excluding the option ##x=c##.
 
  • Love
Likes member 731016
Back
Top