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

Bobbie Comstock 2022-01-03 Answered
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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Expert Answer

Cheryl King
Answered 2022-01-04 Author has 4767 answers
Hint:
\(\displaystyle{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{\sin{{6}}}{x}}}{{{\sin{{2}}}{x}}}}\cdot\sqrt{{{\sin{{2}}}{x}}}\)
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peterpan7117i
Answered 2022-01-05 Author has 2478 answers
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:
\(\displaystyle\lim_{{{x}\to{0}^{+}}}{\frac{{{\sin{{6}}}{x}}}{{\sqrt{{{\sin{{2}}}{x}}}}}}={\frac{{{6}\sqrt{{0}}{C}^{{5}}-{20}\cdot{0}^{{2}}\sqrt{{0}}{C}^{{3}}+{6}\cdot{0}^{{4}}\sqrt{{0}}{C}}}{{\sqrt{{{2}{C}}}}}}={0}\)
0
Vasquez
Answered 2022-01-08 Author has 8850 answers

Hint
\(\frac{\sin 6x}{\sqrt{\sin 2x}}=\frac{6x \frac{\sin 6x}{6x}}{\sqrt{2x} \sqrt{\frac{\sin 2x}{2x}}}\)

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