Proving the limit laws
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. \(\square\)
This work is licensed under a Creative Commons Attribution 4.0 International License.