$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$

Question

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

  $\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$

I've tried converting it into different functions like $\cos(\pi/2-2x)$ or multiplying by the inverse function and so on, but it keep getting back to $0/0$.

I'm sure that the limit does in fact exist because using L'Hôpital's rule it is fairly easy to prove it, but I can't use it because the course I'm attending still hasn't defined the rule's properties, but I couldn't find a different way aside from that.

Answer 1

Hint: $\dfrac{\sin(6x)}{\sqrt{\sin 2x}}= \dfrac{\sin 6x}{\sin 2x}\cdot \sqrt{\sin 2x}$

Answer 2

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

$$\sin\left(6x\right) = 6 \sin x \cos^5 x - 20 \sin^3 x \cos^3 x + 6 \sin^5 x \cos x$$

$$\sqrt{\sin\left(2x\right)} = \sqrt{2 \sin\left(x\right) \cos\left(x\right)}$$

For shorthand, let me write: $$\sin\left(x\right) = S \hspace{2.54cm} \cos\left(x\right) = C$$

So your ratio is: \begin{align*}\lim_{x \rightarrow 0^+} \frac{\sin\left(6x\right)}{\sqrt{\sin\left(2x\right)}} &= \frac{6SC^5 - 20S^3C^3 + 6S^5C}{\sqrt{2SC}} \\ & = \frac{6\sqrt{S}C^5 - 20S^2\sqrt{S}C^3 + 6S^4\sqrt{S}C}{\sqrt{2C}}\end{align*}

Finally, we have a form where the denominator won't blow up. Note that $$\lim_{x \rightarrow 0^+} \sin\left(x\right) = 0 \hspace{2.54cm} \lim_{x \rightarrow 0^+} \cos\left(x\right) = 1$$ to seal the deal: $$\lim_{x \rightarrow 0^+} \frac{\sin\left(6x\right)}{\sqrt{\sin\left(2x\right)}} = \frac{6\sqrt{0}C^5 - 20\cdot 0^2\sqrt{0}C^3 + 6\cdot 0^4\sqrt{0}C}{\sqrt{2C}} = 0$$

Answer 3

Hint

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