Expert: Alon Mandes - 12/11/2009

Question
QUESTION: Dear sir,

I have difficulty understanding the formal proofs to the Limit of Sum Law and Limit of Multiplication Law. From reading, I understand that the rigorous proof involves using triangle inequality, which I understand by itself, but not when it is used in combination with deltas and epsilon.
I think additional attention may be needed on the choice of the values of delta and epsilon because I am unable to grasp the reasons behind the parameters assigned to each of them.
Most of all, I have difficulty understanding mathematical prose, so if possible, explaining the proofs in plain English will be very helpful.

Thank you for your help.

ANSWER: The definition of
Lim f(x) = L
x->a
Given any real %26 small number ε>0  , there exists another real %26 small number δ>0  so that
if 0<|x-a|<δ , then |f(x)-L|<ε .
We want to prove that :
If Lim f(x)=L %26 Lim g(x)=M then
  x->a         x->a
Lim F(x) + Lim g(x) = L+M
x->a       x->a   

Let's prove it step by step :
What do we know ?
We know that :
1. Given any real %26 small number ε1>0  , there exists another real %26 small number δ1>0  so that
  if 0<|x-a|<δ1 , then |f(x)-L|<ε1 .
2. Given any real %26 small number ε2>0  , there exists another real %26 small number δ2>0  so that
  if 0<|x-a|<δ2 , then |f(x)-M|<ε2 .
Let's examine the form |f(x)+g(x)-L-m| , it can also be written as |f(x)-L + g(x)-M| .
If we recall the  triangle inequality : |x+y|≤|x|+|y|. This is essential, because we can use
it in our case : We can claim that |[f(x)-L] + [g(x)-M]|≤|f(x)-L|+|g(x)-M| . Now, we already
know that |f(x)-L|<ε1 for every 0<|x-a|<δ1  %26 |g(x)-M|<ε2 for every 0<|x-a|<δ2 . Thus, we
might claim that |[f(x)-L] + [g(x)-M]|≤ε1+ε1 for every x that satisfy : |x-a|<δ1 %26 |x-a|<δ2 .
So If we choose ε to be  ε=ε1+ε1 %26 δ=min{δ1,δ2} then surely the following statement is true :
Given any real %26 small number ε>0  , there exists another real %26 small number δ>0  so that
if 0<|x-a|<δ , then |f(x)+g(x)-(L+M)|<ε . %26 is the definition of the limit of f+g . Q.E.D

Here's a numerical example to illustrate the above :
If |f(x)-5|<0.001 for every |x-3|<0.02 %26 if |g(x)-10|<0.01 for every |x-3|<0.01 then for every
|x-3|<0.02 ("which is this case is also <0.01") we get |f+g-15|≤|f-5|+|g-10|=0.001+0.01

Alon.
Alon.





---------- FOLLOW-UP ----------

QUESTION: Thank you for the clear explanation.
I have one lingering doubt, however.
Does this show that Lim [f(x)+g(x)]=Lim f(x)+Lim g(x)? If so, how?
         x->a          x->a     x->a
Or to put it in another way, how do I define Lim [f(x)+g(x)] in terms of delta and epsilon and why is it so?

Sorry for the inconvenience and thank you for your help.

Answer
As I mentioned above , the definition of the term : Lim  f(x) = L is :
         x->a
for every real and small number ε>0  , there exists another real and small number δ>0  so that
if 0<|x-a|<δ , then |f(x)-L|<ε .
You may look at ε AND δ as an  accuracy parameters.
So, yes When I proved in ε-δ terms the accumulation of the limits, then automatically I proved
it also in Lim term .

Alon.