# If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a}g(x)$ exists?

This is a question from Spivak's Calculus. Question statement:

If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a} g(x)$ exists?

I have not been able to find a counterexample, so it seems that it exists. Here is my attempt at proof, and I'm mostly interested in validity of my proof.

Following is given:

$\forall \epsilon, \exists \delta_1: 0 < |x-a| < \delta_1 \Rightarrow |f(x) - L_1| < \epsilon$

$\forall \epsilon, \exists \delta_2: 0 < |x-a| < \delta_2 \Rightarrow |f(x) + g(x) - L_2| < \epsilon$
Let $\delta_2 \le \delta_1$. If greater $\delta_2$ works, then smaller one will of course work, and if smaller one works than that inequality follows from that. Let $L_2 - L_1 = L_3$, that is $L_2 = L_3 + L_1$. Then we substitute:

$|f(x) + g(x) - L_1 - L_3| < \epsilon$
$|(g(x) - L_3) - (L_1 - f(x))| < \epsilon$
And by the inequality $|a-b| \ge |a| - |b|$:
$|g(x) - L_3| - |L_1 - f(x)| \le |(g(x) - L_3) - (L_1 - f(x))| < \epsilon$
$|g(x) - L_3| - |f(x) - L_1| < \epsilon$
$|g(x) - L_3| < \epsilon + |f(x) - L_1|$ Since $\delta_2 \le \delta_1$, we can always choose epsilon greater than $|f(x) - L_1|$ so we can make substitution:
$|g(x) - L_3| < 2\epsilon$
And we can make $2\epsilon$ as small as we wish. This completes the proof.

Thank you for any help!

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Do you already know that if $\lim\limits_{x\to c} F(x)$ exists and $\lim\limits_{x\to c} G(x)$ exists, then $\lim\limits_{x\to c}(F-G)$ exists? If so, set $F=f+g$ and $G=f$. –  Arturo Magidin Mar 11 '11 at 22:02
@Lovre: Your proof is valid; it can be streamlined a bit if you simply start from $|g(x)-L_3|$ and use the regular triangle inequality:\begin{align*}|g(x)-L_3| &= |g(x)-L_3 + f(x)-f(x)+L_2-L_2|\\\ &= |(g(x)+f(x)-L_2)-(f(x)-(L_2-L_3))|\\\ &\leq |g(x)+f(x)-L_2|+|f(x)-L_1|.\end{align*} –  Arturo Magidin Mar 11 '11 at 22:06
Ooooh. Thanks very much, very neat way... Didn't occur to me. Though I'm still wondering if my proof is correct, because I want to know whether I have good grasp on concepts or I have a gap in knowledge somewhere. –  user5501 Mar 11 '11 at 22:07
Your proof is OK, except for the formulation "we can always choose epsilon greater than" -- you can't choose $\epsilon$, since you have to find $\delta$ for given $\epsilon$ -- but the proof works nevertheless -- a better way to express what you're doing would be to say that you choose $\delta=\min(\delta_1,\delta_2)$; then both inequalities with $\epsilon$ hold for $\lvert x-a\rvert<\delta$. Also, it would be clearer and more precise if you chose $\epsilon/2$ for $L1$ and $L2$ at the outset and then ended up with $\epsilon$ at the end, since you want to show $\lvert g(x)-L_3\rvert<\epsilon$. –  joriki Mar 11 '11 at 22:14
@Lovre: Concerning your desire to know whether you have a good grasp on the concepts: I think so, but it seems you may not have completely internalized the $\epsilon/\delta$ paradigm. I find the easiest way to understand it is as a two-player game: Player A comes up with an $\epsilon$, and Player B has to come up with a $\delta$ to make the inequalities come out right. The limit is proved if Player B can always win. I suggest you look at your proof and try to see how you slightly confused the roles of the $\epsilon$ being given (which you called $2\epsilon$)... –  joriki Mar 11 '11 at 22:21

So it can be marked off as answered...

• Do you already know that if $\lim\limits_{x\to c}F(x)$ exists and $\lim\limits_{x\to c} G(x)$ exists, then $\lim\limits_{x\to c}\bigl(F(x)-G(x)\bigr)$ exists? If so, set $F(x) = f(x)+g(x)$ and $G(x) = f(x)$ to get the desired result.

• The proof is (essentially) valid; it can be streamlined a bit if you simply start from $|g(x)-L_3|$ and use the triangle inequality: \begin{align*}|g(x)-L_3| &= |g(x)-L_3 + f(x)-f(x)+L_2-L_2|\\ &= \left|\Bigl(g(x)+f(x)-L_2\Bigr)-\Bigl(f(x)-(L_2-L_3)\Bigr)\right|\\ &\leq \left|\Bigl(g(x)+f(x)\bigr)-L_2\right|+\Bigl|f(x)-L_1\Bigr|\\ &\lt 2\epsilon. \end{align*}

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I think your demonstration is correct. Maybe it is "more elegant" to take $\epsilon/2$ in the following sentences:

∀ϵ,∃δ1:0<|x−a|<δ1⇒$|f(x)−L_1|$<ϵ/2

∀ϵ,∃δ2:0<|x−a|<δ2⇒$|f(x)+g(x)−L_2|$<ϵ/2,

in a way that $|g(x)-L_3|<\epsilon$, given that $|x-a| < \delta_2$.

Note that this is the same as the following: If $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} h(x) = M$, then $\lim_{x\to a} [f(x)+h(x)] = L+M$.

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Thanks for the answer. :) But isn't that last thing proved for "if such limits exist", and we can't infer that it wouldn't exist if one of the limits didn't exist? –  user5501 Mar 11 '11 at 22:17
@Lovre: Writing $\lim_{x\to a}f(x) = L$ means "the limit exists, and is equal to $L$". Perhaps you are confusing it with$$\lim_{x\to a}(f+g) = \lim_{x\to a}f + \lim_{x\to a}g,$$which only holds "if all limits exist" (or more precisely, "if at least two of the limits exist", in which case all three will). –  Arturo Magidin Mar 11 '11 at 22:27
@Arturo: Actually, I think that was my point. What I wanted to say was that I wanted to prove that, as you say "if at least two of the limits exist" then all three will, where f+g and f are given to exist, while Ronaldo in last sentence outlined case when f and g exist. –  user5501 Mar 11 '11 at 22:32
In the last sentence I wrote, we assume the two limits $lim_{x\to a} f(x)$ and $lim_{x\to a} h(x)$ exist, and the conclude that the limit of the sum exists. The connection to your question is made writing $h(x) = - [f(x) + g(x)]$ (whose limit exists). –  Ronaldo Mar 11 '11 at 22:43