Can I write: $\lim \limits _{x \to x_0} (f(x) + g(x)) = \lim \limits _{x \to x_0} f(x) + \lim \limits _{x \to x_0} g(x)$? If $\lim \limits _{x \to x_0} (f(x) + g(x))$ exists, can I write: $\lim \limits _{x \to x_0} (f(x) + g(x)) = \lim \limits _{x \to x_0} f(x) + \lim \limits _{x \to x_0} g(x)$ ? I mean, to write this do I have to know that the other limits exist? Because they tell me that $\lim \limits _{x \to x_0} f(x)$ exists and I want to prove that $\lim \limits _{x \to x_0} g(x)$ also exists by writing: $$\lim \limits _{x \to x_0} (f(x) + g(x)) = \lim \limits _{x \to x_0} f(x) + \lim \limits _{x \to x_0} g(x)$$
$$\lim \limits _{x \to x_0} g(x) = \lim \limits _{x \to x_0} (f(x) + g(x)) - \lim \limits _{x \to x_0} f(x)$$
Then because the other limits exist, the theorem says that the limit of $g$ also exists... Is it correct to prove & write that way? 
 A: Observe that if $f(x)=\frac{1}{x}$ and $g(x)=-\frac{1}{x}$, then 
$$
\lim_{x\rightarrow 0}f(x)\text{ DNE}
$$
and 
$$
\lim_{x\rightarrow 0}g(x)\text{ DNE}
$$
However, $f(x)+g(x)=0$ for $x\not=0$.
Therefore, 
$$
\lim_{x\rightarrow 0}f(x)+g(x)=0.
$$
Therefore, even though $\lim_{x\rightarrow 0}f(x)+g(x)$ exists, the individual limits on $f(x)$ and $g(x)$ do not exist.
This contradicts the first line of the question (the OP later adds the condition that $\lim_{x\rightarrow x_0}f(x)$ exists).
A: One of the first facts about limits to know is that if for functions $f$ and $g$ the limits $\lim_{x\to x_0}f(x)$ and $\lim_{x\to x_0}g(x)$ exist, then
$$\lim_{x\to x_0}(f(x)+g(x))=\lim_{x\to x_0}f(x)+\lim_{x\to x_0}g(x).$$
If you haven't already, you should try to prove this for yourself.
From this it follows that if $\lim_{x\to x_0}(f(x)+g(x))$ and $\lim_{x\to x_0}f(x)$ exist, then
$$\lim_{x\to x_0}g(x)=\lim_{x\to x_0}(f(x)+g(x))-\lim_{x\to x_0}f(x).$$
Simply rewrite the above as $a(x):=f(x)+g(x)$ and $b(x)=-f(x)$, so that $a(x)+b(x)=g(x)$, and apply the fact I mentioned above.
A: No. In general, it is possible that $\lim_{x \to x_0} (f(x) + g(x))$ exists while both limits $\lim_{x \to x_0} f(x)$, $\lim_{x \to x_0} g(x)$ doesn't exist (think about 
$$\lim_{x \to \infty} 0 = \lim_{x \to \infty} (x - x) = \lim_{x \to \infty} x + \lim_{x \to \infty} (-x)$$
so it doesn't make any sense to write
$$ \lim_{x \to x_0} (f(x) + g(x)) = \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x) $$
without knowing in advance that both limits $\lim_{x \to x_0} f(x)$, $\lim_{x \to x_0} g(x)$ exist. If you know in advance that both limits in the right hand side exist, then the equality indeed holds.
In your case, you don't have to write it as you can use your second equation and note that
$$ \lim_{x \to x_0} g(x) = \lim_{x \to x_0} (f(x) + g(x) - f(x)) = \lim_{x \to x_0} (f(x) + g(x)) - \lim_{x \to x_0} f(x) $$
and since you know that both limits $\lim_{x \to x_0} (f(x) + g(x))$ and $\lim_{x \to x_0} f(x)$ exist, you can conclude that $\lim_{x \to x_0} g(x)$ exists and equal to $\lim_{x \to x_0} (f(x) + g(x)) - \lim_{x \to x_0} f(x)$.
A: I prefer to write
$$g(x)=(f(x)+g(x))+(-f(x))$$
or $g(x)=(f(x)+g(x))-f(x)$ and then quote standard theorems about the limit of a sum or difference. That keeps the logic of the argument simpler. 
