## Thomas' Calculus 13th Edition

We want to prove $\underset{x\to 4}{\lim}\left(9-x\right)=5$. First, we must find a $\delta > 0$ such that for any $\epsilon > 0$, we have $\left|9-x-5\right| \lt \epsilon$. Note that $$\left|9-x-5\right|=\left|4-x\right|=\left|x-4\right|.$$ So $$\left|9-x-5\right|\lt \epsilon \Leftrightarrow \left|x-4\right|\lt \epsilon.$$ Thus for any $\epsilon > 0$, we let $\delta=\epsilon$. Here is our proof. Let $\epsilon > 0$, and let $\delta=\epsilon$. Then for all $x$ with $0 \lt \left|x-4\right| \lt \delta$, we have $$\left|x-4\right| \lt \epsilon \\ \left|4-x\right|\lt \epsilon \\ \left|9-x-5\right| \lt \epsilon.$$ Hence $\underset{x\to 4}{\lim}\left(9-x\right)=5$.