Epsilon-Delta Definition of Limits

  • Thread starter Thread starter lizielou09
  • Start date Start date
  • Tags Tags
    Definition Limits
Click For Summary
SUMMARY

The discussion focuses on proving the limit as x approaches 0 of f(x) equals the limit as x approaches a of f(x-a) using the epsilon-delta definition of limits. The user establishes that if the absolute difference between the two limits A and B is less than any positive epsilon, then A must equal B. The proof hinges on demonstrating that the conditions for the epsilon-delta definition for f(x) at 0 imply the same for f(x-a) at a, thereby confirming the equality of the two limits.

PREREQUISITES
  • Epsilon-delta definition of limits
  • Understanding of limit notation and properties
  • Basic knowledge of functions and continuity
  • Familiarity with mathematical proofs
NEXT STEPS
  • Study the epsilon-delta definition of limits in detail
  • Explore examples of limits involving transformations of functions
  • Learn about continuity and its implications on limits
  • Practice proving limits using formal epsilon-delta arguments
USEFUL FOR

Mathematics students, educators, and anyone interested in understanding the formal foundations of calculus and limit proofs.

lizielou09
Messages
3
Reaction score
0
Using the epsilon-delta definition of limit, prove that the limit as x approaches 0 of f(x) equals the limit as x approaches a of f(x-a).


I let the limit as x approaches 0 of f(x) equal A and the limit as x approaches a of f(x-a) equal B. If the absolute value of (A-B) is less than epsilon for all positive epsilon, then A=B. If A does not equal B, then let epsilon equal the absolute value of (A-B) divided by 2. Where do I go from here?
 
Physics news on Phys.org
Since you are give that limit as x --> 0 of f(x) = A you have the following statement:

Given ε >0 there is a δ > 0 such that |f(x) - A| < ε when 0 < |x| < δ

And you are trying to prove limit as x --> a of f(x-a) = A which is:

Given ε >0 there is a δ > 0 such that |f(x-a) - A| < ε when 0 < |x-a| < δ

If you think about it carefully you should be able to see how the first statement implies the second one.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
16
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
19
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K