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?

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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.

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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.

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Thank you for any help!