Epsilon Delta Limits: Finding \delta

[jrjack]

Homework Statement

Suppose |f(x)-5|&lt;0.1 when 0<x<5.
Find all values \delta&gt;0 such that |f(x)-5|&lt;0.1 whenever 0&lt;|x-2|&lt;\delta

Homework Equations

The Attempt at a Solution

I know that 0&lt;|x-2|&lt;\delta
2-\delta&lt;x&lt;2+\delta
\delta=2
but how does this part of the equation help me find delta?
|f(x)-5|&lt;0.1
4.9&lt;f(x)&lt;5.1

I don't undestand it's use in this problem, if the other part gave me \delta=2

 

[estro]

Try translating the problem into plain English, then it will be something like:
How far can I move from x=2 so that my function won't be too far from 5, while 0.1 is already too far...=)

BTW, how is this related to limits? [here you need to find appropriate \delta]

 

[HallsofIvy]

Homework Statement

Suppose |f(x)-5|&lt;0.1 when 0<x<5.
Find all values \delta&gt;0 such that |f(x)-5|&lt;0.1 whenever 0&lt;|x-2|&lt;\delta

Homework Equations

The Attempt at a Solution

I know that 0&lt;|x-2|&lt;\delta
2-\delta&lt;x&lt;2+\delta
\delta=2
but how does this part of the equation help me find delta?
|f(x)-5|&lt;0.1
4.9&lt;f(x)&lt;5.1

You don't need this at all. You are given that |f(x)- 5|&lt; 0.1 if 0< x< 5 and you want "|f(x)- 5|< 0.1 if 2-\delta&lt; x&lt; 2+ \delta" so the "f" part is the same in both hypothesis and conclusion. Focus on the other part

I don't undestand it's use in this problem, if the other part gave me \delta=2

Ignore f completely. What value of \delta will guarantee that if 2-\delta&lt; x&lt; 2+ \delta then 0&lt; x&lt; 5?

 

[jrjack]

Thanks, I have been watching the Kahn Academey and you tube videos and I'm starting to grasp this. I take this course on-line through a community college and the instructors lesson was a power point slide with no sound...it was lacking a lot of description and any explanation.

The videos, on the other hand, were very helpful, so is advice on here, Thanks.