# Homework Help: Not understanding simple proof

1. Jan 3, 2012

### Miike012

Theorem: If f approaches l and g approaches m neer a then

lim(f+g) = l + m as x approaches a

Proof:

If 0 < | x - a | < delta1, then |f(x) - l| < epsilon,
and If 0 < | x - a | < delta2 then |g(x) - m| < epsilon,

If 0 < | x - a | < delta1, then |f(x) - l| < epsilon / 2,
and If 0 < | x - a | < delta2 then |g(x) - m| < epsilon/2,

Now Let Delta = min(delta1,delta2). If 0 < | x - a | < delta, then 0 < | x - a | < delta1 and
0 < | x - a | < delta2 are both true, so both
|f(x) - l| < epsilon / 2 and |g(x) - m| < epsilon/2 are true and |(f + g)(x) - (l + m)| < epsilon.

I understand what the theorem is saying but when it goes into proving the theorem I dont fully understand the logic. I have never had any formal class on proofs and I am trying to read through this book but it takes me 2-3 hours to get through one chapter then at the end I still don't fully understand the material. My question is, will taking a discrete mathematics class, if im right this is kind of like a intro to math proofs and logic, should it make understanding proofs like this and harder ones easier to understand? If not this class, are there any other classes that I can take?
I'm thinking if I dont understand something as "easy" as this proof then Im not ment for math, even though I average 95-97 on math tests, but that obviously meens nothing.

2. Jan 3, 2012

### Poopsilon

You seem to have essentially gotten it, although it's a bit sloppy. You have let ε/2 > 0 and based on what you were given, have found a single δ such that if |x - a|<δ than both |f(x) - l|<ε/2 and |g(x) - m|<ε/2. And thus since the less-than-sign respects addition you added the inequalities to obtain |f(x) - l|+|g(x) - m|< ε/2 + ε/2 and then used the triangle quality to conclude that |f(x) + g(x) - (l+m)|< ε.

An intro to proofs class may help, although epsilon-delta proofs are notoriously convoluted at first, it's not just you. Unfortunately they are the very definition, in a metric space at least, of what it means for a function to 'respect all sequences in the domain which converge to some point'. If that point is in your domain as well then this just becomes continuity (bit of a subtlety).

It might be beneficial to draw a graph of a continuous curve, and observe how no matter how close together you make your horizontal epsilon bars around any point on the curve, you can always put two vertical delta bars close enough together such that they cross only the part of the curve within the epsilon bars. Now make your graph have a jump discontinuity and put some really close together epsilon bars around the discontinuous point, and notice how no matter how close together you make your delta bars you always catch a bit of each side of the curve around the empty whole in your graph where the point used to be.. hope that made sense

Last edited: Jan 3, 2012
3. Jan 3, 2012

### Staff: Mentor

This looks a lot like the basic theorem of calculus. Maybe start with that proof to understand as x approaches some value. I think you have to think of the f and g functions as mappings
and not as formulas that you plug values into.

As x gets closer to a then there's a delta1 value that just larger than x-a ok
We know that there's always a related epsilon that is just larger than the f-l

Similarly for the g with same epsilon but with a different delta named delta2

Your mission is to proof the two deltas are the same and when you do you can state that:

That the limit of a sum of functions is the same as the sum of the limits.

By the way the limit proof was very tough for me too. It's the reason calculus works.
For many years after its developments mathematicians didn't fully trust the
results until concepts like smoothness and continuity were well defined.

Notice

Last edited: Jan 3, 2012
4. Jan 3, 2012

### Miike012

What I was actually saying is that... the proof that I posted above was from the book, and I dont fully understand the steps in the proof, although I have an idea why they made them I still dont fully understand all the steps.

Last edited: Jan 3, 2012
5. Jan 3, 2012

### Poopsilon

Ah I see, well the khan academy has some epsilon delta explanations on youtube if you are still not clear on just a straight epsilon delta proof of a limit.

Beyond that maybe you could be more specific about what exactly you don't understand in the steps.

6. Jan 3, 2012

### SammyS

Staff Emeritus
Even though this is the book's proof, I agree with Poopsilon, it is sloppy, especially for a textbook.

The first few lines could be set aside as being explanatory -- although not part of the proof:
If 0 < | x - a | < delta1, then |f(x) - l| < epsilon,
and If 0 < | x - a | < delta2 then |g(x) - m| < epsilon.

If 0 < | x - a | < delta1, then |f(x) - l| < epsilon / 2,
and If 0 < | x - a | < delta2 then |g(x) - m| < epsilon/2.​

I'll draw up a clearer (I hope) proof.

Theorem: If $\displaystyle \lim_{x\to a}\ f(x)=L$ and $\displaystyle \lim_{x\to a}\ g(x)=M$, then $\displaystyle \lim_{x\to a}\ \left(f(x)+g(x)\right)=L+M$.

Explanatory aside:
What does $\displaystyle \lim_{x\to a}\ f(x)=L$ mean?

It means that if we are given any positive number, ε+, no matter how small, then we know that there exists a positive number, δ+, so that for any x that satisfies, $\displaystyle 0<|x-a|<\delta^+\,,$ we are assured that $\displaystyle |f(x)-L|<\varepsilon^+\,.$​

Proof:
Let ε > 0.
We need to show: There is a δ such that if 0 < |x-a| < δ, then |f(x)+g(x)-(L+M)| < ε.
Of course, ε/2 > 0, so we know that $\displaystyle \lim_{x\to a}\ f(x)=L$ implies that there is a number δ1 > 0 such that whenever $\displaystyle 0<|x-a|<\delta_1\,,$ then $\displaystyle |f(x)-L|<\frac{\varepsilon}{2}\,.$

Similarly, $\displaystyle \lim_{x\to a}\ g(x)=M$ implies that there is a number δ2 > 0 such that whenever $\displaystyle 0<|x-a|<\delta_2\,,$ then $\displaystyle |g(x)-M|<\frac{\varepsilon}{2}\,.$

Let $\displaystyle \delta=\min(\delta_1,\,\delta_2)\,.$ This tells us that δ ≤ δ1 and δ ≤ δ2 .

Now we will only consider values of x which are a distance less that δ from a but x ≠ a .

Let $\displaystyle 0<|x-a|<\delta\,.$ So we also know that, 0 < |x-a| < δ1 and 0 < |x-a| < δ2

Then

$\displaystyle |f(x)+g(x)-(L+M)| = |f(x)-L+g(x)-M|$

$\displaystyle \le |f(x)-L|+|g(x)-M|$ (by the triangle inequality)

$<\displaystyle \frac{\varepsilon}{2}+\frac{\varepsilon}{2}$

$=\varepsilon$​
QED.

The text in red is comment, not part of the proof.

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