Find limit of $\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}$, if it exists I'm practising solving limits and the one I'm currently struggling with is the following: $$\ell =\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}$$
What I've done:


*

*Since this is an obvious $0/0$ , I tried using de L'Hospital's Rule consecutively only to see both the numerator and the denominator grow so much in size that each couldn't fit in one row.
$$
\begin{align}
l
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}\sin(4x)}}\\
& = \lim_{x→0}{{200\sin^{199}(x)\cos(x)}\over{x^{198}\left(199\sin\left(4x\right)+4x\cos\left(4x\right)\right)}}\\
& = \lim_{x→0}{{39800\cos^2\left(x\right)\sin^{198}\left(x\right)-200\sin^{200}\left(x\right)}\over{x^{198}\left(800\cos\left(4x\right)-16x\sin\left(4x\right)\right)+198x^{197}\left(199\sin\left(4x\right)+4x\cos\left(4x\right)\right)}}
\end{align}
$$

*Another solution I tried was through manipulation and the use of trigonometric identities and formulae but to no avail. I tried substituting:


*

*$\color{red}{\sin(4x)}$ with $\color{blue}{4\sin(x)\cos(x) - 8\sin(3x)\cos(x)}$ and then

*$\color{red}{8\sin(3x)\cos(x)}$ with $\color{blue}{4\sin(4x)+4\sin(2x)}$.
$$
\begin{align}
l
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}\sin(4x)}}\\
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}(4\sin(x)\cos(x) - 8\sin(3x)\cos(x))}}\\
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}(4\sin(x)\cos(x) - 4\sin(4x)+4\sin(2x))}}\\
\end{align}
$$



No matter what I try, the limit remains $0/0$.
Question:
Does the above limit exist? If so, what I path should I follow to work out a solution?

Since $\sin x= x+o(x)$ we have, $$\frac{\sin^{200}x}{x^{199}\sin(4x)}= \frac{x^{200}+o(x^{200})}{x^{199}(4x+o(x))}=\frac{x^{200}+o(x^{200})}{4x^{200}+o(x^{200})}=\frac{1+o(1)}{4+o(1)} =\to\frac{1}{4}$$
