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?
 A: Use $\lim_{x \to 0} \frac{\sin x}{x} = 1$:
$$\lim_{x \to 0} \frac{x^3}{\tan^3 2x} = \lim_{x\to 0} \left( \frac{(2x)^3}{\sin^3 2x} \cdot \frac{\cos^3 2x}{8} \right) \stackrel{[1 \cdot \frac{1}{8}]}{=} \frac{1}{8}$$
A: $$\lim_{x \to 0} \frac{x^3}{\tan (2x)^3}=\lim_{x \to 0} \frac{x^3}{(2x)^3}=\frac18$$
A: $$ \lim_{x \rightarrow 0 }\frac{x^3}{\tan^3 (2x) }  =  
\lim_{x \rightarrow 0 } \left ( \frac{2x}{\tan (2x) } \right )^3 \frac{1}{2^3} = \frac{1}{2^3}$$
$$ $$
A: $\tan x =x +o(x)$ then $\tan (2x) \sim 2x$ , then : $\frac{x^3}{\tan^3 x} \sim \frac{x^3}{8x^3} \sim \frac 18$
A: Another idea using $\,\,\displaystyle{\frac{\sin x}{x}\underset{x\to 0}{\longrightarrow}1\,,\,\cos kx\underset{x\to 0}\longrightarrow 1\,\,(k=\text{a constant})\,\,,\,\sin 2x=2\sin x\cos x}$:
$$\frac{x^3}{\tan^3 2x}=\frac{x^3}{\frac{\sin^32x}{\cos^32x}}=\cos^32x\frac{x^3}{\left(2\sin x\cos x\right)^3}=\frac{1}{8}\frac{\cos^32x}{\cos^3x}\left(\frac{x}{\sin x}\right)^3\underset{x\to 0}\longrightarrow \frac{1}{8}\cdot\frac{1}{1}\cdot 1^3=\frac{1}{8}$$
A: We may resort to  $\sin(x)<x<\tan(x),\space 0< x <\frac{\pi}{2}$ and solve it elementarily. By Squeeze's theorem we get that:
$$\lim_{x\rightarrow0}\frac{x^3 \cos^3(2x)}{{(2x)}^3}\leq \lim_{x\rightarrow0}\frac{x^3}{\tan^3(2x)}\leq \lim_{x\rightarrow0}\frac{x^3}{(2x)^3}$$
Therefore, taking also into account the symmetry the limit is $\frac{1}{8}$.  
The proof is complete.
A: $\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$=$\lim_{x\rightarrow 0} \frac{x^3}{\frac{\sin^3(2x)}{\cos^3(2x)}}$=$\lim_{x\rightarrow 0} \frac{x^3\cos^3(2x)}{\sin^3(2x)}$=$\lim_{x\rightarrow 0}\frac{x\cdot x\cdot x \cdot\cos^3(2x)}{\sin(2x)\cdot
\sin(2x)\cdot\sin(2x)}$=$\lim_{x\rightarrow 0}\frac{2x\cdot 2x\cdot 2x \cdot \frac{1}{8} \cos^3(2x)}{\sin(2x)\cdot\sin(2x)\cdot\sin(2x)}$=$|\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$\Rightarrow$ $\lim_{x\rightarrow 0}\frac{x}{\sin x}=1$  $\Rightarrow$ $\lim_{x\rightarrow 0}\frac{2x}{\sin 2x}=1$|=$\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$1\cdot 1\cdot 1\cdot \frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}{\cos[\lim_{x\rightarrow 0}(2x)]^3}$=$\frac{1}{8}\cos0$=$\frac{1}{8}\cdot 1$=$\frac{1}{8}$
