epsilon Delta approach in Proving $\lim_{x \to 8} \sqrt[3]x=2$ using epsilon Delta approach  Prove that $$\lim_{x \to 8} \sqrt[3]x=2$$
Given $\epsilon \gt 0$ we need to find $\delta=f(\epsilon)$ such that $$|x-a| \lt \delta$$ $\implies$
$$|f(x)-L| \lt \epsilon$$  So
$$|\sqrt[3]x -2| \lt \epsilon$$  $\implies$
$$2-\epsilon \lt \sqrt[3]x \lt 2+\epsilon$$  Or
$$(2-\epsilon)^3 \lt x \lt (2+\epsilon)^3$$  Now using
$(a-b)^3$ and neglecting higher powers of $\epsilon$ we get
$$8-12 \epsilon \lt x \lt 8+12\epsilon$$
So
$$-12 \epsilon \lt x-8 \lt 12 \epsilon$$
So $$\delta=12 \epsilon$$
But my book answer is $$\delta =min \left\{8-(2-\epsilon)^3, 8-(2+\epsilon)^3\right\}$$
can any one clarify this
 A: You have shown
$$|\sqrt[3] x-2|<\epsilon \iff (2-\epsilon)^3<x<(2+\epsilon)^3$$
This is equivalent to$$8-(2+\epsilon)^3<8-x<8-(2-\epsilon)^3$$
which is implied by
$$|x-8|<\delta = \min\{(2+\epsilon)^3-8,8-(2-\epsilon)^3\}$$
A: First of all note that such questions don't have a unique answer and hence trying to match the answer from the book does not make any sense. Second and most importantly such questions are not an exercise to solve inequalities via algebraic manipulation.
For any given $\epsilon>0 $ you need to ensure that the inequality $$|\sqrt[3]{x}-2|<\epsilon\tag{1}$$ holds by constraining the values of $x$ in a certain manner. The constraint on the values of $x$ needs to be of the form $0<|x-8|<\delta$ where the $\delta$ has to be chosen so that the inequality $(1)$ is ensured.
The correct strategy here is not to solve the inequation $(1)$ to find all values of $x$ for which the inequality $(1)$ holds, but rather to find a simple expression in $x$, say $g(x) $, such that inequality $$|\sqrt[3]{x}-2|\leq g(x) \tag{2}$$ holds and then ensure that inequality $$g(x) <\epsilon\tag{3}$$ holds. Ensuring that inequalities $(2)$ and $(3)$ hold automatically ensures that inequality $(1)$ holds and naturally ensuring these two inequalities will require some constraints on values of $x$ and we will combine these constraints into a single one to get something of the form $0<|x-8|<\delta$.
The crux of the problem is thus to find a suitable and simple expression $g(x) $ which satisfies $(2)$. And we would want it to be as simple as possible. Note that $$|\sqrt[3]{x}-2|=\left|\frac{x-8}{\sqrt[3]{x^{2}}+2\sqrt[3]{x}+4}\right|$$ Now we can observe that if $x>1$ then the denominator $$\sqrt[3]{x^{2}}+2\sqrt[3]{x}+4>1+2+4=7$$ and hence we have $$|\sqrt[3]{x}-2|<\frac{|x-8|}{7}$$ if $x>1$. We have thus found our $g(x) $ as $g(x) = |x-8|/7$. And we also have a constraint on $x$ namely $x>1$ so that this $g(x) $ works as expected. We can put this constraint as $0<|x-8|<7$. And then we need to ensure $g(x) =|x-8|/7<\epsilon$ which holds if we have the constraint on $x$ in the form of $0<|x-8|<7\epsilon$. We now combine the two constrains on $x$ as $$0<|x-8|<\min(7,7\epsilon)$$ and it is clear that we can take $\delta=\min(7,7\epsilon)$.

You can see that the above does not involve any messy algebra and gives a simple and clear answer (much simpler than the complicated answer given in your book). 
