Prove $\lim_{x\to 4}\sqrt{x^3}=8$ $\lim_{x\to 4}\sqrt{x^3}=8$
My attempt:
Given $\epsilon>0$, choose $\delta=\min(1, \frac{\epsilon}{61})$
$|\sqrt{x^3}-8|=|\frac{(x-4)(x^2+4x+16)}{\sqrt{x^3}+8}|=|x-4||x^2+4x+16||\frac{1}{\sqrt{x^3}+8}|$
If $|x-4|<1$ then $-5<x<5$
So, $|x^2+4x+16|<61$
And $|\frac{1}{\sqrt{x^3}+8}|<1$
So, $|\sqrt{x^3}-8|<61|x-4|<61\delta=\epsilon$
Is this valid? 

The proof is good. You could avoid some steps by factoring
$$
\sqrt{x^3}-8=(\sqrt{x}-2)(x+2\sqrt{x}+4)
$$
and observe that, if $|x-4|<1$, then $|x+2\sqrt{x}+4|<5+2\sqrt{5}+4<15$ and you just need to take $|\sqrt{x}-2|<\frac{\varepsilon}{15}$; since
$$
|\sqrt{x}-2|=\left|\frac{x-4}{\sqrt{x}+2}\right|<|x-4|
$$
when $x>3$, you can take $\delta=\min(1,\varepsilon/15)$.
