# Real Analysis Limits

## 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.

## 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.

Dick
Science Advisor
Homework Helper

## 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.

## 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?

Dick
Science Advisor
Homework Helper
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?