## Calculus 8th Edition

$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that the following is valid: $($if $0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon)$ ------------- Given any $\epsilon > 0$, we want to find a $\delta > 0$ such that $0 < | x-10| < \delta \ \ \Rightarrow\ \ |3-\displaystyle \frac{4}{5}x-(-5)| < \epsilon$. Analyzing $|3-\displaystyle \frac{4}{5}x-(-5)| < \epsilon,\ \qquad(*)$ we write equivalent inequalities: $|8-\displaystyle \frac{4}{5}x| < \epsilon$, factor out $\displaystyle \frac{4}{5}$ on the LHS, as we aim for $| x-10|$ ...$8=\displaystyle \frac{40}{5}$ $|\displaystyle \frac{4}{5}||10-x| < \epsilon$ $|\displaystyle \frac{4}{5}||x-10| < \epsilon\qquad /\times\frac{5}{4}$ $|x-10| < \displaystyle \frac{5}{4}\epsilon$ Remember that this is equivalent to (*). So, for any given $\epsilon > 0$, we take $\displaystyle \delta=\frac{5}{4}\epsilon$ and $0 < |x-10| < \delta \ \ \Rightarrow\ \ |3-\displaystyle \frac{4}{5}x-(-5)| < \epsilon$, (the implication is one of two that comprise the equivalence) which, by the definition of a limit, means that. $\displaystyle \lim_{x\rightarrow 10}(3-\frac{4}{5}x)=-5$