--- title: Proving the limit laws tags: math, analysis --- **Addition.** Let $E\subset \mathbf{R}$ be a subset of the real numbers, $f, g : E \to \mathbf{R}$ be real functions, and $L_1, L_2\in \mathbf{R}$ be real numbers. Now suppose $\lim_{x\to a} f(x) = L_1$ and $\lim_{x\to a} g(x) = L_2$. We would like to show that $\lim_{x\to a} (f(x)+g(x)) = L_1+L_2$. Since $\lim_{x\to a} f(x) = L_1$ and $\lim_{x\to a} g(x) = L_2$, we know that $$\forall \epsilon >0 \exists \delta >0 \forall x (0 < |x-a|<\delta \to |f(x)-L_1|<\epsilon)$$ and $$\forall \epsilon >0 \exists \delta >0 \forall x (0 < |x-a|<\delta \to |g(x)-L_2|<\epsilon)$$ Now let $\epsilon > 0$. We need to find $\delta > 0$ such that $$0<|x-a|<\delta \to |(f(x)+g(x))-(L_1+L_2)|<\epsilon$$ But \begin{align} |f(x)+g(x)-(L_1+L_2)| &= |f(x)-L_1 + g(x)-L_2| \\ &\leq |f(x)-L_1| + |g(x)-L_2| \end{align} by the triangle inequality. So if we can show $|f(x)-L_1| + |g(x)-L_2|<\epsilon$, then our result follows immediately. But, $\frac{\epsilon}{2}$ being a positive real number, we have $\delta_1,\delta_2 > 0$ such that $$0 < |x-a|<\delta_1 \to |f(x)-L_1|<\frac{\epsilon}{2} \\ 0 < |x-a|<\delta_2 \to |f(x)-L_2|<\frac{\epsilon}{2}$$ So if we let $\delta := \min(\delta_1,\delta_2)$, then $|f(x)-L_1|$ and $|g(x)-L_2|$ are both less than $\frac{\epsilon}{2}$, so $$|f(x)-L_1| + |g(x)-L_2|< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ which completes the proof. $\square$