derivatives - Find the limit $quadlim_{x to 0}frac{x^3}{tan^3(2x)}$

I'm trying to find the limit $$\quad\lim_{x \to 0}\frac{x^3}{\tan^3(2x)}$$ but I'm at a loss.

I've tried expanding $\tan^3(2x)$ using $\tan2x = \frac{\sin2x}{\cos2x}$, and then using double-angle formulas to expand that, but that did not yield anything helpful from what I could tell.

---

This is a question from an end-of-chapter exercise. Up and til this chapter this book has only dealt with introductory-type limits and consequently introduced the derivative and differentiation rules. That being said, I'm not supposed to apply some fancy technique that I've not been introduced to thus far.

Any suggestions?

Answer

Hint: $$\lim_{x \to 0}\frac{x^3}{\sin^3(2x)}=\lim_{x \to 0}\frac{1}{8}\cdot\frac{(2x)^3}{\sin^3(2x)}=\frac{1}{8}\cdot1$$