# Proving the limits of the sum of two functions is equal to the sum of the limits

I am new to proving in math so I want to know if this informal proof of limits is possible:

Theorem: If $\lim_{x \to a}f(x)=A$ and $\lim_{x \to a}g(x) = B$, then $$\lim_{x \to a}[f(x)+g(x)]=A+B$$

$\lim_{x \to a}[f(x) + g(x)] = A + B$ is the same as $\lim_{x\to a}[(f(x)-A) + (g(x) - B)]=0$.

Also, $$0 \le |(f(x)-A) + (g(x) - B)| \le |f(x)-A| + |g(x) - B|.$$ Since $\lim_{x\to a} f(x)-A=0$ and $\lim_{x\to a}g(x)-B=0$, we know $$0 \le \lim_{x\to a}|(f(x)-A) + (g(x) - B)| \le 0.$$ So, $\lim_{x\to a}[(f(x)-A) + (g(x) - B)] = 0.\ \square$

If my proof is wrong what is wrong with it and how can I correct it? Also, why is the proof in Apostol II on page 248 of limits correct? Why didn't he use limits ($\epsilon,\delta$) as he did in Apostol I on page 132? Also, why did he assume that the limits A and B are 0 then say that proves it for all cases?

This is Apostol's proof(Theorem 8.1) in vector valued functions of vector variables. Is this proof also valid for real valued functions of vector variables(scalar fields) ? Why didn't he use $\epsilon,\delta$ ?

Your proof starts with a good technique namely reducing the problem to a simpler case of it. Here you have $\lim_{x \to a}\{f(x) - A\} = 0$ and $\lim_{x \to a}\{g(x) - B\} = 0$ and you have to establish that $\lim_{x \to a}\left\{\{f(x) - A\} + \{g(x) - B\}\right\} = 0$

However you write that "we know that $$0 \leq \lim_{x \to a}|(f(x) - A) + (g(x) - B)| \leq 0$$ This step is not obvious from $\lim_{x \to a}\{f(x) - A\} = 0$ and $\lim_{x \to a}\{g(x) - B\} = 0$. At some point you do need to use $\epsilon , \delta$ arguments.

It would be better however to proceed to directly with $f(x), g(x)$ instead of $f(x) - A$ and $g(x) - B$. Try to use the definition of limits and there should not be any problem.

About Apostol's proof in Calculus II, can you show it here (by typing or more easily by attaching an image)? Then it would be possible to comment on it.

Update: After looking at Apostol proof (image attached) it is clear that he is using the already established theorems on limits of real valued functions (in Apostol 1 page 132) and he is using them very smartly to prove theorems related to vector valued functions.

In Apostol 1, he does make use of $\epsilon, \delta$, but he proves the theorems under the assumption that $A = B = 0$ and says that this is sufficient to handle the general case. This means for example, that Apostol proves using $\epsilon, \delta$ (in Apostol 1) the following theorem:

If $\lim_{x \to a}f(x) = 0, \lim_{x \to a}g(x) = 0$ then $\lim_{x \to a}\{f(x) + g(x)\} = 0$

Assuming that you have understood this proof from Apostol 1, your approch to prove the result below is OK.

If $\lim_{x \to a}f(x) = A, \lim_{x \to a}g(x) = B$ then $\lim_{x \to a}\{f(x) + g(x)\} = A + B$

But you can't ignore the basic result when $A = B = 0$ which was proved on Page 132 of Apostol 1.

Further Update (based on OP's comments): How does $A = B = 0$ version imply the general version? I have to say that you have answered exactly this part correctly in your question. Only thing was that you did not know that you were assuming the already established special case of $A = B = 0$ and that's why I objected that your statement "we know that ...." is not obvious.

• I deduced this step from the fact that 0<=|[f(x)-A] + [g(x)-B]|<=|[f(x)-A| + |g(x)-B| from the triangle inequality. And since we have lim of f(x)-A = 0 and lim of g(x)-B = 0 so 0 <= lim |[f(x)-A] + [g(x)-B]| <= 0 thus it is 0. – Nameless Nov 9 '13 at 11:49
• The triangle inequality is fine but not the inequality of type $0 \leq \lim_{x \to a}\cdots \leq 0$. I also see the proof from Apostol. In this proof the author (he is by the way a very good author) assumes the following: if $f(x), g(x)$ are non negative and tend to $0$ as $x \to a$ then their sum $f(x) + g(x)$ also tends to $0$ as $x \to a$. This is something which you need to prove via $\epsilon, \delta$. Perhaps Apostol wants readers to prove this simple stuff on their own. I will put such a proof in my updated answer if you wish. – Paramanand Singh Nov 9 '13 at 14:26
• From further looking at Apostol I see that the functions $f(x), g(x)$ he talks about are not real numbers, but rather vectors (he uses $\textbf{bold}$ letters for vectors) and the notation he uses as $\left\|a\right\|$ is the norm (or length) of vector $a$ and this norm is a real number. So I believe Apostol has proved the basic facts on limits of real valued functions in some earlier chapter. You should consult that. – Paramanand Singh Nov 9 '13 at 14:32
• Apostol proved it on real valued functions of real variables NOT real valued functions of vector variables. Do you mean that Apostol used the proved statement from real valued functions of REAL variable to prove the vector valued function of vector variable? Also, when proving it for real valued functions of real variables he used $\epsilon,\delta$, that is why am curious why he didn't use $\epsilon,\delta$ in this proof. – Nameless Nov 9 '13 at 14:38
• Thank you for the updated answer it made things a lot clearer. Also(this is another question), why did Apostol use A=B=0, aren't proofs supposed to be for the $"general"$ case ? To make my question clearer, how does the proof of the case A=B=0 imply the proof of the $"general"$ case ? – Nameless Nov 9 '13 at 14:56

Your proof looks fine.

I don't have Apostol so I can't look at his proof.

To see how the case for $0$ implies all cases, just do exactly what you did.

Take $h(x) = f(x) - A$ and $k(x) = g(x) - B$.

Then $h(x), k(x) \rightarrow 0$ and if we prove the theorem for $h, k$ we also get it for $f$ and $g$ by the first line of your proof.

• I have put Apostol's II page 248 proof. Can you comment on it ? Also, I can't understand how my proof is okay although I didn't use $\epsilon,\delta$. – Nameless Nov 9 '13 at 11:55