limits - Without using Taylor expansion, how do I find the value of K if $lim_{x rightarrow 0}frac{x^2 sin(Kx)}{x - sin x} =1$

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

Here K is a constant whose value I want to find,

I got it by writing the series expansion of $\sin(\theta)$, but couldn't by L'hospital rule or standard limits,

This is what I tried:

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

$$\implies \lim_{x \rightarrow 0}\frac{x \sin(Kx)}{1 - \frac{\sin x} x} =1$$

$$\implies \lim_{x \rightarrow 0}\frac{Kx^2 \frac{\sin(Kx)}{Kx}}{1 - \frac{\sin x} x} =1$$

So this gives a 0 in the denominator which doesn't help, L'hospital gives a huge mess, applying it once more didn't help

The answer given is K =1/6.