## 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 $0 < | x-1| < \delta \ \ \Rightarrow\ \ |\displaystyle \frac{2+4x}{3}-2| < \epsilon$. Analyzing $|\displaystyle \frac{2+4x}{3}-2| < \epsilon$, we write equivalent inequalities: $|\displaystyle \frac{2+4x}{3}-2| < \epsilon$ $|\displaystyle \frac{4x-4}{3}| < \epsilon$ $|\displaystyle \frac{4}{3}||x-1| < \epsilon \displaystyle \qquad/\times\frac{3}{4}$ $|x-1| < \displaystyle \frac{3}{4}\epsilon$. The first and last inequalities are equivalent. $\displaystyle \frac{3}{4}\epsilon$ is a positive nonnegative number, and we take it for our $\delta$. So (equivalence being double implication), $\displaystyle \delta=\frac{3}{4}\epsilon$, \$0