Basic limit question ( f(x) = x^2 )

[ltjrpliskin]

f(x) = x^2 and we want to show that as x -> a, f(x) -> a^2.

So, we must ensure that |x^2 - a^2| < eps.

The book (Calculus by Spivak) starts of by bounding |x - a| < 1 which leads us to
|x + a| < 2|a| + 1.

This is the part where I get lost:
I can understand that we require |x - a| < eps/ (2|a| + 1),
but
why do we choose |x - a| < min(1, eps/ (2|a| + 1)) ?

And what happens if min(1, eps/ (2|a| + 1)) = 1 ?

Thank you very much for your help in advance! :)

 

[tiny-tim]

hi ltjrpliskin!

(try using the X² button just above the Reply box )

|x² - a²| = |x - a||x + a|,

so we need |x + a| to be bounded …

any bound would do, but it seems most sensible to use 1

 

[ltjrpliskin]

hi ltjrpliskin!

(try using the X² button just above the Reply box )

|x² - a²| = |x - a||x + a|,

so we need |x + a| to be bounded …

any bound would do, but it seems most sensible to use 1

Okay, but can you help me understand why we chose the minimum of 1 and eps/ (2|a| + 1)? :)

 

[tiny-tim]

we want |x - a||x + a| to be small …

|x - a| < 1 keeps |x + a| small, in fact < (2|a| + 1)

|x - a| < ε/ (2|a| + 1), keeps |x - a| small,

and both together keep the whole thing < ε ​

so we need both conditions … it won't work if |x - a| can be unbounded

 

[ltjrpliskin]

we want |x - a||x + a| to be small …

|x - a| < 1 keeps |x + a| small, in fact < (2|a| + 1)

|x - a| < ε/ (2|a| + 1), keeps |x - a| small,

and both together keep the whole thing < ε ​

so we need both conditions … it won't work if |x - a| can be unbounded

Ahh! Okay, thank you very much :)
That was much simpler than expected haha :)