## Calculus 8th Edition

By definition , $\displaystyle \lim_{x\rightarrow 2}(14-5x)=4$ is true if for any $\epsilon > 0$, there is a $\delta > 0$ such that if $0 < |x-2| < \delta$, then $|(14-5x)-4| < \epsilon$. Analyzing the last expression, $|(14-5x)-4| < \epsilon$ $|-5x+10| < \epsilon$ $|-5||x-2| < \epsilon$ $|x-2| < \displaystyle \frac{\epsilon}{5}$. So, for any given $\epsilon$, if we choose $\displaystyle \delta=\frac{\epsilon}{5}$,we will have $0 < |x-2| < \delta \Rightarrow |(14-5x)-4|$