Solution
= Solution
Rationalize the difference of <square roots> before taking the <limit>. For positive $x$, divide the resulting numerator and denominator by $x^2$:
$$
\frac{8x^2+2x}{\bigl(\sqrt{x^3+10x^2}+\sqrt{x^3+2x^2-2x}\bigr)\sqrt{6x+3}}
=\frac{8+2/x}{\bigl(\sqrt{1+10/x}+\sqrt{1+2/x-2/x^2}\bigr)\sqrt{6+3/x}}.
$$
Each factor now has a finite nonzero denominator <limit>, giving
$$
\boxed{\frac4{\sqrt6}=\frac{2\sqrt6}{3}.}
$$