---
layout: default
title: Proving the limit laws
comments: "yes"
disqus-id: ee9b153b71076aa4d5c8e683bb33f2adb0b78f0c
math: "yes"
license: "CC-BY"
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. $\square$