- #1
njama
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Homework Statement
[TEX][|x^2-9|< \epsilon [/TEX] if [tex]0 < |x-3| < \delta[/tex]
Homework Equations
The Attempt at a Solution
Hello!
I know how to solve this task but I have certain things that I do not understand.
Solution.
[TEX][|x^2-9|< \epsilon [/TEX] if [tex]0 < |x-3| < \delta[/tex]
Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.
[tex]|x+3||x-3|< \epsilon[/tex] if [tex]0<|x-3|<\delta[/tex]
Using the triangle inequality:
[tex]|x+3|=|(x-3)+6| \leq |x-3| + 6[/tex]
By multiplying the statement above with |x-3| (which is positive) we got:
[tex]|x+3||x-3| \leq (|x-3| +6)|x-3| [/tex]
Now clearly [tex]6+|x-3|< 6 + \delta[/tex] and [tex]|x-3|<\delta[/tex]
By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:
[tex](6+|x-3|)|x-3|< (6 + \delta)\delta[/tex]
We can conclude that:
[tex]|x+3||x-3| < (6 + \delta)\delta[/tex] if [tex]0 < |x-3| < \delta[/tex]
Now isn't [tex]\epsilon = (6 + \delta)\delta[/tex] ??
Why in my book they choose [tex](6 + \delta)\delta \leq \epsilon[/tex] ?
And why they choose [tex]\delta \geq 1 [/tex] and with that restriction
[tex](6 + \delta)\delta \leq 7\delta[/tex] and [tex]7\delta \leq \epsilon[/tex].
Finally I don't understand why [tex]\delta = min(e/7,1)[/tex].
Could somebody possibly explain?
Thanks a lot.