# I'm trying to prove and compute the limit of this

I'm trying to prove and compute the limit of this function.
$$\displaystyle\lim_{{{x}\to{0}^{+}}}{\frac{{{\sin{{\left({6}{x}\right)}}}}}{{\sqrt{{{\sin{{\left({2}{x}\right)}}}}}}}}$$
I've tried converting it into different functions like $$\displaystyle{\cos{{\left({\frac{{\pi}}{{{2}}}}−{2}{x}\right)}}}$$ or multiplying by the inverse function and so on, but it keep getting back to 0/0.

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Cheryl King
Hint:
$$\displaystyle{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{\sin{{6}}}{x}}}{{{\sin{{2}}}{x}}}}\cdot\sqrt{{{\sin{{2}}}{x}}}$$
###### Not exactly what you’re looking for?
peterpan7117i
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:
$$\displaystyle{\sin{{\left({x}\right)}}}={S}\ \ \ \ \ {\cos{{\left({x}\right)}}}={C}$$
$$\displaystyle\lim_{{{x}\to{0}^{+}}}{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{6}{S}{C}^{{5}}-{20}{S}^{{3}}{C}^{{3}}+{6}{S}^{{5}}{C}}}{{\sqrt{{{2}{S}{C}}}}}}$$
$$\displaystyle={\frac{{{6}\sqrt{{{S}}}{C}^{{5}}-{20}{S}^{{2}}\sqrt{{{S}}}{C}^{{3}}+{6}{S}^{{4}}\sqrt{{{S}}}{C}}}{{\sqrt{{{2}{C}}}}}}$$
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:
$$\displaystyle\lim_{{{x}\to{0}^{+}}}{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{6}\sqrt{{0}}{C}^{{5}}-{20}\cdot{0}^{{2}}\sqrt{{0}}{C}^{{3}}+{6}\cdot{0}^{{4}}\sqrt{{0}}{C}}}{{\sqrt{{{2}{C}}}}}}={0}$$
Vasquez

Hint
$$\frac{\sin 6x}{\sqrt{\sin 2x}}=\frac{6x \frac{\sin 6x}{6x}}{\sqrt{2x} \sqrt{\frac{\sin 2x}{2x}}}$$