Proving the limit laws

Last modification date: 2016-05-01
Generated on: 2021-06-09

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 ${lim}_{x \to a} f (x) = L_{1}$ 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 $\forall ϵ > 0 \exists δ > 0 \forall x (0 < | x - a | < δ \to | f (x) - L_{1} | < ϵ)$ and $\forall ϵ > 0 \exists δ > 0 \forall x (0 < | x - a | < δ \to | g (x) - L_{2} | < ϵ)$ Now let $ϵ > 0$ . We need to find $δ > 0$ such that $0 < | x - a | < δ \to | (f (x) + g (x)) - (L_{1} + L_{2}) | < ϵ$ But $\begin{matrix} | f (x) + g (x) - (L_{1} + L_{2}) | & = | f (x) - L_{1} + g (x) - L_{2} | \leq | f (x) - L_{1} | + | g (x) - L_{2} | \end{matrix}$ by the triangle inequality. So if we can show $| f (x) - L_{1} | + | g (x) - L_{2} | < ϵ$ , then our result follows immediately. But, $\frac{ϵ}{2}$ being a positive real number, we have $δ_{1}, δ_{2} > 0$ such that $0 < | x - a | < δ_{1} \to | f (x) - L_{1} | < \frac{ϵ}{2} 0 < | x - a | < δ_{2} \to | f (x) - L_{2} | < \frac{ϵ}{2}$ So if we let $δ := min (δ_{1}, δ_{2})$ , then $| f (x) - L_{1} |$ and $| g (x) - L_{2} |$ are both less than $\frac{ϵ}{2}$ , so $| f (x) - L_{1} | + | g (x) - L_{2} | < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ$ which completes the proof. $□$