Answer
please see step-by-step
Work Step by Step
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$ )
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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
$\displaystyle \lim_{x\rightarrow-6^{+}}\sqrt[8]{6+x}=0$ .
