Proving $\lim_{x\to x_0}f(x)+g(x)=\lim_{x\to x_0}f(x)+\lim_{x\to x_0}g(x)$ 
Suppose that $\lim_{x\to x_0}f(x)=L$ and $\lim_{x\to x_0}g(x)=M$. I was attempting to prove that $\lim_{x\to x_0}f(x)+g(x)=L+M$.

The following is my attempt at a proof.
Proof: We have to show that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all those values of $x$ for which $|x-x_0|<\delta$ we have that $|f(x)+g(x)-(L+M)|<\varepsilon$.
Because $\lim_{x\to x_0}f(x)=L$ and $\lim_{x\to x_0}g(x)=M$, we know that $|f(x)-L|<\frac\varepsilon 2$ and $|g(x)-M|<\frac \varepsilon 2$ for all values of $x$ for which $|x-x_0|<\delta$. Adding the two inequalities we get that $|f(x)-L|+|g(x)-M|<\varepsilon $. Using the triangle inequality we get that $|f(x)+g(x)-(L+M)|< \varepsilon$ for all those values of $x$ for which $|x-x_0|<\delta$.
In other words, $\lim_{x\to x_0}f(x)+g(x)=L+M$.
Is the proof correct?

You need to be a bit more careful, as the $\delta$'s for the two limits might not be the same. Indeed if $\delta_1$ is the one for $f$, and $\delta_2$ is the one for $g$, then set $\delta=\min\{\delta_1,\delta_2\}$ and your argument works.
