Expert: Socrates - 8/27/2009

Question
Dear Sir,

I am currently studying the precise definition of limits. As to the Proof of Sum Law, I am unsure why 0 < |x - a| < δ1 will result in |f(x) - L| < ε/2, similarly 0 < |x - a| < δ2 will result in |g(x) - M| < ε/2 . What is the rationale for this? Why use ε/2?

Also, when doing questions on precise limits, why do I need to verify my δ? For example:

lim x->3 (3x) = 9

Eventually I can "solve" the equation such that δ = ε/3. Now that I have a value of δ that corresponds with ε, why do I still need to go back and "verify" by writing the statements all over again?

I am not a Math major, and I am struggling with my current Math module, so I really need all the help I can get. I really appreciate you taking the time out to help me with this.

Thanks!

Answer
I assume you are trying to show that the limit of the sum of two functions ,f and g , will be the sum of the limits , L and M.

Because you are given that the limit for f(x) as x=> a is L , you know that there is a δ1 such that |f(x) - L| < ε/2 when 0 < |x - a| < δ1

Similarly , you know that there is a δ2 such that |g(x) - M| < ε/2 when 0 < |x - a| < δ2 because you know that the limit for g(x) as x=> a is M

You use ε/2 because you are going to combine the two inequalities |f(x) - L| < ε/2 and
|g(x) - M| < ε/2  by adding corresponding sides and you want to end up with ε on the right side.

|f(x) - L| + |g(x) - M| < ε/2 + ε/2 = ε

use the triangle inequality


|(f(x)+g(x)) - (L+M)| < |f(x) - L| + |g(x) - M| < ε

so

|(f(x)+g(x)) - (L+M)| < ε


when 0 < |x - a| < δ where δ = minimum of δ1 and δ2 because choosing the minimum

makes both |f(x) - L| < ε/2  and |g(x) - M| < ε/2 true simultaneously.


As to why you must verify your value for δ = ε/3 , I am not sure , maybe it is just to make sure you haven't made any mistakes in arithmetic and get more practice with the definition of the limit.