Proving the limit laws

Addition. Let $E\subset R$ be a subset of the real numbers, $f,g:E\to R$ be real functions, and $L_1,L_2\in R$ be real numbers. Now suppose ${lim}_{x\to a}f\left(x\right)=L_1$ and ${lim}_{x\to a}g\left(x\right)=L_2$. We would like to show that ${lim}_{x\to a}\left(f\right(x)+g(x\left)\right)=L_1+L_2$.

Since ${lim}_{x\to a}f\left(x\right)=L_1$ and ${lim}_{x\to a}g\left(x\right)=L_2$, we know that $$\forall ϵ>0\exists \delta >0\forall x(0<|x-a|<\delta \to |f(x)-L_1|<ϵ)$$ and $$\forall ϵ>0\exists \delta >0\forall x(0<|x-a|<\delta \to |g(x)-L_2|<ϵ)$$ Now let $ϵ>0$. We need to find $\delta >0$ such that $$0<|x-a|<\delta \to |\left(f\right(x)+g(x\left)\right)-(L_1+L_2)|<ϵ$$ But $$\begin{array}{cc}|f\left(x\right)+g\left(x\right)-(L_1+L_2)|& =|f\left(x\right)-L_1+g\left(x\right)-L_2|\\ & \le |f\left(x\right)-L_1|+|g\left(x\right)-L_2|\end{array}$$ by the triangle inequality. So if we can show $|f\left(x\right)-L_1|+|g\left(x\right)-L_2|<ϵ$, then our result follows immediately. But, $\frac{ϵ}{2}$ being a positive real number, we have ${\delta }_1,{\delta }_2>0$ such that $$0<|x-a|<{\delta }_1\to |f\left(x\right)-L_1|<\frac{ϵ}{2}0<|x-a|<{\delta }_2\to |f\left(x\right)-L_2|<\frac{ϵ}{2}$$ So if we let $\delta :=min({\delta }_1,{\delta }_2)$, then $|f\left(x\right)-L_1|$ and $|g\left(x\right)-L_2|$ are both less than $\frac{ϵ}{2}$, so $$|f\left(x\right)-L_1|+|g\left(x\right)-L_2|<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$$ which completes the proof. $\square$