- #1
-Dragoon-
- 309
- 7
Find the largest δ that "works".
Find the largest δ that "works" for the given ϵ:
[itex]\displaystyle \lim_{x\to1}2x = 2; ϵ = 0.1[/itex]
N/A
Given ϵ > 0, then:
if [itex]0 < |x - 1| < δ[/itex] then [itex]|2x - 2| < ϵ[/itex]
ϵ = 0.1, so, [itex]|2x - 2| < 0.1[/itex]
Now to establish the connection:
[itex] |2x - 2| => |2 (x - 1)| => |2||x - 1| => 2|x - 1|[/itex]
Therefore: [itex] 2|x - 1| < ϵ => |x - 1| < \frac{ϵ}{2} => |x - 1| <\frac{0.1}{2} =>|x - 1| < 0.05.[/itex] The largest value that "works" for δ is 0.05 since if:
[itex] 0 < |x - 1| < 0.05[/itex] then [itex]|2x - 2| < 0.1[/itex]
But, in my textbook, the answer is 2ϵ = 0.2 as the largest value that "works" for δ. So, I just wanted to know what I did wrong in my calculations as the book only shows the solution and not the work. Thanks in advance.
Homework Statement
Find the largest δ that "works" for the given ϵ:
[itex]\displaystyle \lim_{x\to1}2x = 2; ϵ = 0.1[/itex]
Homework Equations
N/A
The Attempt at a Solution
Given ϵ > 0, then:
if [itex]0 < |x - 1| < δ[/itex] then [itex]|2x - 2| < ϵ[/itex]
ϵ = 0.1, so, [itex]|2x - 2| < 0.1[/itex]
Now to establish the connection:
[itex] |2x - 2| => |2 (x - 1)| => |2||x - 1| => 2|x - 1|[/itex]
Therefore: [itex] 2|x - 1| < ϵ => |x - 1| < \frac{ϵ}{2} => |x - 1| <\frac{0.1}{2} =>|x - 1| < 0.05.[/itex] The largest value that "works" for δ is 0.05 since if:
[itex] 0 < |x - 1| < 0.05[/itex] then [itex]|2x - 2| < 0.1[/itex]
But, in my textbook, the answer is 2ϵ = 0.2 as the largest value that "works" for δ. So, I just wanted to know what I did wrong in my calculations as the book only shows the solution and not the work. Thanks in advance.