Epsilon Delta Limits: Finding \delta

In summary, the problem is asking for the value of \delta that will guarantee that if 2-\delta< x< 2+ \delta then 0< x< 5. This can be translated into plain English as "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?" The solution is to focus on the other part of the equation and ignore the "f" part completely. The value of \delta=2 will guarantee that if 2-\delta< x< 2+ \delta then 0< x< 5.
  • #1
jrjack
111
0

Homework Statement


Suppose [itex]|f(x)-5|<0.1[/itex] when 0<x<5.
Find all values [itex]\delta>0[/itex] such that [itex]|f(x)-5|<0.1[/itex] whenever [itex]0<|x-2|<\delta[/itex]


Homework Equations





The Attempt at a Solution


I know that [tex]0<|x-2|<\delta[/tex]
[tex] 2-\delta<x<2+\delta[/tex]
[tex]\delta=2[/tex]
but how does this part of the equation help me find delta?
[tex]|f(x)-5|<0.1[/tex]
[tex]4.9<f(x)<5.1[/tex]

I don't undestand it's use in this problem, if the other part gave me [itex]\delta=2[/itex]
 
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  • #2
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 [itex]\delta[/itex]]
 
Last edited:
  • #3
jrjack said:

Homework Statement


Suppose [itex]|f(x)-5|<0.1[/itex] when 0<x<5.
Find all values [itex]\delta>0[/itex] such that [itex]|f(x)-5|<0.1[/itex] whenever [itex]0<|x-2|<\delta[/itex]


Homework Equations





The Attempt at a Solution


I know that [tex]0<|x-2|<\delta[/tex]
[tex] 2-\delta<x<2+\delta[/tex]
[tex]\delta=2[/tex]
but how does this part of the equation help me find delta?
[tex]|f(x)-5|<0.1[/tex]
[tex]4.9<f(x)<5.1[/tex]
You don't need this at all. You are given that [itex]|f(x)- 5|< 0.1[/itex] if 0< x< 5 and you want "|f(x)- 5|< 0.1 if [itex]2-\delta< x< 2+ \delta[/itex]" 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 [itex]\delta=2[/itex]

Ignore f completely. What value of [itex]\delta[/itex] will guarantee that if [itex]2-\delta< x< 2+ \delta[/itex] then [itex]0< x< 5[/itex]?
 
  • #4
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.
 

1. What is the purpose of finding "delta" in an epsilon delta limit?

The purpose of finding "delta" in an epsilon delta limit is to determine the minimum distance between the input value and the limit that guarantees the output value will be within a certain range (epsilon) from the limit.

2. How do you find "delta" in an epsilon delta limit?

To find "delta" in an epsilon delta limit, you must first determine the desired range (epsilon) from the limit. Then, you must manipulate the limit expression to isolate "delta" as the subject of the inequality. Finally, you can use algebraic techniques to solve for "delta".

3. What is the significance of "delta" in an epsilon delta limit?

The significance of "delta" in an epsilon delta limit is that it represents the maximum distance between the input value and the limit within which the output value will be guaranteed to fall. It is a crucial part of the definition of a limit and allows for precise mathematical descriptions of continuous functions.

4. Can "delta" ever be negative in an epsilon delta limit?

No, "delta" cannot be negative in an epsilon delta limit. "Delta" represents a distance, and distance cannot be negative. Additionally, the definition of a limit specifies that "delta" must be greater than zero.

5. What happens if the value of "delta" is too small or too large in an epsilon delta limit?

If the value of "delta" is too small, it may not encompass the desired range (epsilon) from the limit, resulting in an incorrect limit. If the value of "delta" is too large, it may encompass a larger range than necessary, but the limit will still be correct. In general, it is best to choose the smallest possible value of "delta" that satisfies the limit definition to ensure the most precise result.

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