You can do this with regular old trig identities. Note that:

\(\displaystyle{\sin{{6}}}{x}={6}{\sin{{x}}}{{\cos}^{{5}}{x}}-{20}{{\sin}^{{3}}{x}}{{\cos}^{{3}}{x}}+{6}{{\sin}^{{5}}{x}}{\cos{{x}}}\)

\(\displaystyle\sqrt{{{\sin{{2}}}{x}}}=\sqrt{{{2}{\sin{{x}}}{\cos{{x}}}}}\)

For shorthand, let me write:

\(\displaystyle{\sin{{\left({x}\right)}}}={S}\ \ \ \ \ {\cos{{\left({x}\right)}}}={C}\)

So your ratio is:

\(\displaystyle\lim_{{{x}\to{0}^{+}}}{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{6}{S}{C}^{{5}}-{20}{S}^{{3}}{C}^{{3}}+{6}{S}^{{5}}{C}}}{{\sqrt{{{2}{S}{C}}}}}}\)

\(\displaystyle={\frac{{{6}\sqrt{{{S}}}{C}^{{5}}-{20}{S}^{{2}}\sqrt{{{S}}}{C}^{{3}}+{6}{S}^{{4}}\sqrt{{{S}}}{C}}}{{\sqrt{{{2}{C}}}}}}\)

Finally, we have a form where the denominator won't blow up. Note that

\(\displaystyle\lim_{{{x}\to{0}^{+}}}{\sin{{\left({x}\right)}}}={0}\ \ \ \ \ \lim_{{{x}\to{0}^{+}}}{\cos{{\left({x}\right)}}}={1}\)

to seal the deal:

\(\displaystyle\lim_{{{x}\to{0}^{+}}}{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{6}\sqrt{{0}}{C}^{{5}}-{20}\cdot{0}^{{2}}\sqrt{{0}}{C}^{{3}}+{6}\cdot{0}^{{4}}\sqrt{{0}}{C}}}{{\sqrt{{{2}{C}}}}}}={0}\)