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

In summary: 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.If f is an even function, then lim x->0+f(x)=L.
  • #1
MathSquareRoo
26
0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 

1. What does "Proving Even Fct Lim x->0 f(x)=L iff Lim x->0+ f(x)=L" mean?

This statement means that the limit of a function as x approaches 0 from both the positive and negative sides is equal to the overall limit of the function as x approaches 0. In other words, the function's limit exists if and only if the limit from both sides exists and is equal.

2. Why is it important to prove this statement?

Proving this statement is important because it helps us understand the behavior of functions at a specific point and can help us make predictions about the function's behavior as x approaches that point. It also allows us to determine if a function is continuous at a certain point.

3. What techniques can be used to prove this statement?

There are several techniques that can be used to prove this statement, including the epsilon-delta definition of a limit, the squeeze theorem, and algebraic manipulation. The specific method used will depend on the function and its properties.

4. Can this statement be applied to all functions?

Yes, this statement can be applied to all functions as long as they are defined at the point in question and satisfy the conditions for existence of a limit.

5. Does this statement only apply to limits at x=0?

No, this statement can be applied to limits at any point, not just x=0. The only requirement is that the function is defined at that point and the limit exists.

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