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

Question

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

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.

Answer 1

$$\lim_{x \rightarrow 0}\frac{x^2 \sin(Kx)}{x - \sin x} =\lim_{x\to 0}\frac{Kx^3}{x-\sin x}=\lim_{x\to 0}\frac{3Kx^2}{1-\cos x}=\lim_{x\to 0}\frac{6Kx}{\sin x}=6K$$

Answer 2

L'Hopital's rule certainly works: you can apply L'Hopital's rule three times to get the limiting function to be continuous at $x=0$.

The third derivative of the numerator is $$-k(6kx\sin(kx)+(k^2x^2-6)\cos(kx))$$ and the third derivative of the denominator is $\cos(x)$. Hence, after three applications of L'Hopital's rule, we get $$1=\lim_{x\to 0}\frac{-k(6kx\sin(kx)+(k^2x^2-6)\cos(kx))}{\cos(x)}=\frac{-k(0+(0-6)\cdot 1)}{1}=6k$$

I leave it to you to perform all the necessary steps, and in particular you must check that after the first and second applications of L'Hopital's rule, we are still in the "$\frac{0}{0}$" indeterminate form which actually allows us to apply L'Hoptial.