We can use L'Hopital's rule or Taylor series expansion to evaluate the limit.

Method 1: L'Hopital's Rule

$\lim_{x\to 0} \frac{\sin(2x) - 2 \sin x}{x^3}$

Applying L'Hopital's rule once:

= $\lim_{x\to 0} \frac{2\cos(2x) - 2 \cos x}{3x^2}$

Applying L'Hopital's rule again:

= $\lim_{x\to 0} \frac{-4\sin(2x) + 2 \sin x}{6x}$

Applying L'Hopital's rule once more:

= $\lim_{x\to 0} \frac{-8\cos(2x) + 2 \cos x}{6}$

= $\frac{-8(1) + 2(1)}{6} = \frac{-6}{6} = -1$

Method 2: Taylor Series Expansion

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$

$\sin(2x) = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - ... = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} - ...$

$\sin(2x) - 2\sin x = (2x - \frac{8x^3}{6} + ...) - 2(x - \frac{x^3}{6} + ...)$

= $2x - \frac{4x^3}{3} - 2x + \frac{x^3}{3} + O(x^5) = -x^3 + O(x^5)$

$\lim_{x\to 0} \frac{\sin(2x) - 2 \sin x}{x^3} = \lim_{x\to 0} \frac{-x^3}{x^3} = -1$

Method 1: L'Hopital's Rule

Let $L = \lim_{x\to 0} \frac{\sin(2x) - 2 \sin x}{x^3}$

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's rule.

Applying L'Hopital's rule once:

$L = \lim_{x\to 0} \frac{2\cos(2x) - 2 \cos x}{3x^2}$

This is still of the form $\frac{0}{0}$, so we apply L'Hopital's rule again:

$L = \lim_{x\to 0} \frac{-4\sin(2x) + 2 \sin x}{6x}$

This is still of the form $\frac{0}{0}$, so we apply L'Hopital's rule once more:

$L = \lim_{x\to 0} \frac{-8\cos(2x) + 2 \cos x}{6}$

Now, we can directly substitute $x = 0$:

$L = \frac{-8\cos(0) + 2 \cos(0)}{6} = \frac{-8(1) + 2(1)}{6} = \frac{-6}{6} = -1$

Method 2: Taylor Series Expansion

We can use the Taylor series expansions for $\sin x$ and $\sin(2x)$ around $x=0$.

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$

$\sin(2x) = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - ... = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} - ...$

Then, $\sin(2x) - 2\sin x = (2x - \frac{8x^3}{6} + ...) - 2(x - \frac{x^3}{6} + ...)$

$= 2x - \frac{4x^3}{3} - 2x + \frac{x^3}{3} + O(x^5) = -x^3 + O(x^5)$

Therefore, $\lim_{x\to 0} \frac{\sin(2x) - 2 \sin x}{x^3} = \lim_{x\to 0} \frac{-x^3 + O(x^5)}{x^3} = -1$