Calculus, how to find the value of delta in $\epsilon$-$\delta$ limit proof 
Problem: Show that $\lim_{x\to 4}x^3=64$. 

My steps:
$|f(x)-L|<\epsilon$ so $|x^3-64|<\epsilon$
$|x^3|<64+\epsilon \rightarrow 64-\epsilon < x^3 < 64+\epsilon$
$\sqrt[3]{64-\epsilon}<x<\sqrt[3]{64+\epsilon}$
$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$
So $\delta<\sqrt[3]{64+\epsilon}-4$
Let $\epsilon>0$. Pick $\delta$ so that $\delta<\sqrt[3]{64+\epsilon}-4$.
Suppose $0<|x-4|<\delta$. Then $|x-4|<\sqrt[3]{64+\epsilon}-4$.
Thus, $|x|<\sqrt[3]{64+\epsilon}$ and cube to get $|x^3|<64+\epsilon$.
Then $|x^3-64|<\epsilon$ which completes the proof.
But this is what the answer said:
Let $\epsilon>0$. Pick $\delta$ so that $\delta<1$ and $\delta<\frac{\epsilon}{61}$.
Suppose $0<|x−4|<\delta$.
Then $4−\delta<x<4+\delta$.
Cube to get $(4−\delta)^3 <x^3 <(4+\delta)^3$.
Expanding the right-side inequality, we get $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64$      $=64+\epsilon$.
The other inequality is similar.
So my questions are:

*

*How can I get $\delta<\frac{\epsilon}{61}$?

*Why $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64=64+\epsilon$?

*When we need to pick $\delta=\min(1,\frac{\epsilon}{61}$) in this case?

*Does my proof has problems ?
I am just a newbie to calculus... Thanks in advance.

Note that in you proof is ok from here
$$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$$
you should set
$$\delta=min\{4-\sqrt[3]{64-\epsilon},\sqrt[3]{64+\epsilon}-4\}=\sqrt[3]{64+\epsilon}-4$$
and what you find is the optimal value for $\delta$.
Note that by binomial expansion
$$\sqrt[3]{64+\epsilon}-4=4\sqrt[3]{1+\epsilon/64}-4\approx\frac{\epsilon}{48}$$
then for $\epsilon$ small also $\frac{\epsilon}{61}<\frac{\epsilon}{48}$ fulfills the condition.
The other limit is obtained by
$$|x^3-64|<\epsilon\iff |x-4||x^2+4x+16|<\epsilon\iff |x-4|<\frac{\epsilon}{ |x^2+4x+16|}=\delta$$
thus for $0<|x-4|<1 \implies3< x<5 \quad x\ne 4$ the minimum value for $\delta$ is attained for $x=5$ that is
$$ \delta_{min}=\frac{\epsilon}{ |5^2+4\cdot5+16|}=\frac{\epsilon}{61} $$
then the condition is satisfy for $\delta=\min\left(1,\frac{\epsilon}{61}\right)$.
