Find limit of $\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}$, if it exists

Question

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?

Answer 1

Note that by standard limit

$${{\sin^{200}x}\over{x^{199}\sin(4x)}}={{\sin^{200}x}\over{x^{200}}}\cdot{{ 4x}\over{\sin(4x)}}\cdot\frac14\to1^{200}\cdot1\cdot\frac14=\frac14$$

Answer 2

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