Limit finding of an indeterminate form: $\lim\limits_{x\to0} \frac{x^3}{\tan^3(2x)}$ Here is the limit I'm trying to find out:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$$
Since it is an indeterminate form, I simply applied l'Hopital's Rule and I ended up with:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)} = \lim_{x\rightarrow 0}\frac{6\cos^3(2x)}{48\cos^3(2x)} = \frac{6}{48} = 0.125$$
Unfortuntely, as far as I've tried, I haven't been able to solve this limit without using l'Hopital's Rule. Is it possibile to algebrically manipulate the equation so to have a determinate form?

Another idea using $\,\,\displaystyle{\frac{\sin x}{x}\underset{x\to 0}{\longrightarrow}1\,,\,\cos kx\underset{x\to 0}\longrightarrow 1\,\,(k=\text{a constant})\,\,,\,\sin 2x=2\sin x\cos x}$:
$$\frac{x^3}{\tan^3 2x}=\frac{x^3}{\frac{\sin^32x}{\cos^32x}}=\cos^32x\frac{x^3}{\left(2\sin x\cos x\right)^3}=\frac{1}{8}\frac{\cos^32x}{\cos^3x}\left(\frac{x}{\sin x}\right)^3\underset{x\to 0}\longrightarrow \frac{1}{8}\cdot\frac{1}{1}\cdot 1^3=\frac{1}{8}$$
