Theorem: If $\lim_{x\to a}\ f(x)=L$ and $\lim_{x\to a}\ g(x)=M$ then $\lim_{x\to a}(f(x)+g(x))=L+M$.
Proof:

Suppose $\lim_{x\to a}\ f(x)=L$ and $\lim_{x\to a}\ g(x)=M$, and consider any $\epsilon >0$. We must find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)+g(x)-(L+M)|<\epsilon$.

First of all, note that $|f(x)+g(x)-(L+M)|=|f(x)-L+g(x)-M|\le |f(x)-L|+|g(x)-M|$ by the triangle inequality (which states that $|a+b|\le|a|+|b|$).

Note also that $\frac{\epsilon}{2}>0$. Since $\lim_{x\to a}\ f(x)=L$ we can find a $\delta_1>0$ such that if $0<|x-a|<\delta_1$ then $|f(x)-L|<\frac{\epsilon}{2}$, and since $\lim_{x\to a}\ g(x)=M$ we can find a $\delta_2>0$ such that if $0<|x-a|<\delta_2$ then $|g(x)-M|<\frac{\epsilon}{2}$.

Take $\delta=\text{min}\{\delta_1,\delta_2\}$ and assume $0<|x-a|<\delta$.

Then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$, so $|f(x)+g(x)-(L+M)|\le |f(x)-L|+|g(x)-M|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

This proves that $\lim_{x\to a}\ (f(x)+g(x))=L+M$.