Proving Even Fct Lim x->0 f(x)=L iff Lim x->0+ f(x)=L

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SUMMARY

The discussion centers on proving that for an even function f: R->R, the limit as x approaches 0, lim x->0 f(x)=L, holds if and only if the right-hand limit, lim x->0+ f(x)=L, is true. The proof hinges on the property of even functions, where f(x)=f(-x). By applying the definitions of limits and substituting -x for x, the equivalence of the limits is established. The participants emphasize the importance of correctly applying the epsilon-delta definition of limits to clarify the proof structure.

PREREQUISITES
  • Understanding of even functions and their properties
  • Familiarity with limit definitions in calculus
  • Knowledge of epsilon-delta proofs
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the epsilon-delta definition of limits in depth
  • Learn about the properties of even and odd functions
  • Practice proving limits using formal definitions
  • Explore examples of limit proofs involving piecewise functions
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Students studying calculus, particularly those focusing on limit proofs, mathematicians interested in function properties, and educators teaching limit concepts in analysis.

MathSquareRoo
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Homework Statement


Prove that if f: R->R is an even function, then lim x->0 f(x)=L if and only if lim x->0+ f(x)=L.


Homework Equations





The Attempt at a Solution



So far I have:

If f is an even function f(x)=f(-x) for x in domain of f.

Then I am trying to apply the limit definitions, but am unsure of how to write the proof from here.
 
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MathSquareRoo said:

Homework Statement


Prove that if f: R->R is an even function, then lim x->0 f(x)=L if and only if lim x->0+ f(x)=L.


Homework Equations





The Attempt at a Solution



So far I have:

If f is an even function f(x)=f(-x) for x in domain of f.

Then I am trying to apply the limit definitions, but am unsure of how to write the proof from here.

Write down the definition of lim x->0+ f(x)=L. Now change x to -x. Doesn't it look like the definition of lim x->0- f(x)=L once you use that f is even?
 
So lim x->0+ f(x)=L implies there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(x)-Ll<epsilon provided 0<x-a<delta.

Then lim x->0+ f(-x)=L implies that there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(-x)-Ll<epsilon provided 0<x-a<delta.

I have the definitions, but I don't understand the last part of your comment, can you clarify?
 
MathSquareRoo said:
So lim x->0+ f(x)=L implies there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(x)-Ll<epsilon provided 0<x-a<delta.

Then lim x->0+ f(-x)=L implies that there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(-x)-Ll<epsilon provided 0<x-a<delta.

I have the definitions, but I don't understand the last part of your comment, can you clarify?


'a' in your problem is 0. 0<x<delta, is the same as -delta<-x<0. What does the definition of lim x->0- f(x)=L look like?
 
lim x->0- f(x)=L implies that there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(x)-Ll<epsilon provided 0<a-x<delta.

I'm am getting confused with all these definitions though, can you help me organize the argument using the definitions?
 

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