## Calculus 8th Edition

See p.77, Definition of Right-Hand Limit $\displaystyle \lim_{x\rightarrow a^{+}}f(x)=L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that ( $a < x < a+\delta \ \ \Rightarrow\ \ |f(x)-L| < \epsilon$ ) ------------- Given $\epsilon > 0$, we want to find a $\delta > 0$ such that $0 < x-(-6) < 0+\delta \ \ \Rightarrow\ \ |\sqrt[8]{6+x}-0| < \epsilon$. Lokking at the conclusion of the implication, we write equivalent inequalities $|\sqrt[8]{6+x}-0| < \epsilon$ $\sqrt[8]{6+x} < \epsilon$ $6+x < \epsilon^{8}$ ...the LHS can be written as $x+6=x-(-6)$ $x-(-6) < \epsilon^{8}$ ... which leads us to take $\delta=\epsilon^{8}$, Then. $0 < x-(-6) < 0+ \delta \Rightarrow |\sqrt[8]{6+x}-0| < \epsilon$. By the definition of a right-hand limit $\displaystyle \lim_{x\rightarrow-6^{+}}\sqrt[8]{6+x}=0$ .