Looking for other approaches to evaluate $\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}$ Here is my approach:  $$\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}=\lim _{x\to 0^+}\frac{1}{\sqrt{\cos^2x+\cos x+1}}\times\frac{\sin 3x}{\sqrt{1-\cos x}}$$
$$=\lim _{x\to 0^+}\frac1{\sqrt3}\times\frac{\sin 3x}{\sqrt{1-\cos x}}\times\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}=\frac{\sqrt2}{\sqrt3}\times\lim _{x\to 0^+}\frac{\sin 3x}{\sin x}=\sqrt6$$
Can you evaluate $\lim _{x\to 0^+}\dfrac{\sin 3x}{\sqrt{1-\cos ^3x}}$ with other approaches?
 A: I would write something like that,
$$\sin(3x) \sim 3x$$
$$\cos^3(x) = \left(1-\dfrac{x^2}{2}+o(x^2)\right)^3 = 1-3\dfrac{x^2}{2}+o(x^2)$$
Hence,
$$1-\cos^3(x) \sim 3\dfrac{x^2}{2} $$
Hence,
$$\sqrt{1-\cos^3(x)} \sim \sqrt{3/2}x  $$
Finally,
$$\dfrac{\sin(3x)}{\sqrt{1-\cos^3(x)}} \sim \dfrac{3}{\sqrt{3/2}} \sim \sqrt{6}$$
A: You can apply L'Hopital's Rule:$$\lim_{x\to0^+}\frac{\sin^2(3x)}{1-\cos^3(x)}=\lim_{x\to0^+}\frac{6\sin(3x)\cos(3x)}{3\sin(x)\cos^2(x)}.\tag1$$And now you can prove (again by L'Hopital's Rule) that $\lim_{x\to0^+}\frac{\sin(3x)}{\sin(x)}=3$, and it will now follow that $(1)$ is equal to $6$.Edit: Actually, you don't have to use L'Hopital's Rule for a second time in order to compute $\lim_{x\to0^+}\frac{\sin(3x)}{\sin(x)}=3$. It follows from the fact that\begin{align}\sin(3x)&=\sin(x+2x)\\&=\sin(x)\cos(2x)+\cos(x)\sin(2x)\\&=\sin(x)\cos(2x)+2\sin(x)\cos^2(x),\end{align}and that therefore$$\lim_{x\to0^+}\frac{\sin(3x)}{\sin(x)}=\lim_{x\to0^+}\bigl(\cos(2x)+2\cos^2(x)\bigr)=3.$$
