Show that in $\lim_{x\to 0}\frac{x^3\sin(1/x)}{\sin^2 x}$ can't be applied L'Hopital's Rule

I was trying to demonstrate this by showing that one or some of the conditions for the application of L'Hopital's Rule are not met, but it seems to me that the conditions work just fine.

According to my textbook. One of the condition of applicability of L'Hopital's Rule is that:

If:

$1)$ $f$ anf $g$ are differentiable on $(a,b)$; $b-a <\infty$;

$2)$ $\lim_{x\to a+0}f(x) = \lim_{x\to a+0}g(x) = 0$;

$3)$ $g' \neq 0$ on $(a,b)$;

$4)$ $\exists \lim_{x\to a+0}\frac{f'(x)}{g'(x)} \in \mathbb{\overline{R}}$

Then $$\lim_{x\to a+0}\frac{f(x)}{g(x)} = \lim_{x\to a+0}\frac{f'(x)}{g'(x)}$$

My thoughts:

Here $a = 0$, so $(0,b)$ is our open interval.

$1)$ Both the numerator and the denominator are differentiable on this open interval.

$2)$ as $x \to +0 f(x) = g(x)$ (considering the boundeness of $\sin (\frac{1}{x}$)

$3)$ we can choose a $b$ small enough so $g'(x) = \sin (2x) \neq 0$ on this interval (any $b<\pi$ works).

$4)$ on $(0, b)$, $\lim_{x\to a+0}\frac{f'(x)}{g'(x)}$ is indeterminate but I can apply again L'Hopital's Rule.

What am I missing?

$$\lim_{x\to 0}\frac{f'(x)}{g'(x)}=\lim_{x\to 0}\left(\frac{3x^2\sin(1/x)-x\cos(1/x)}{2\sin(x)\cos(x)}\right)$$
fails to exist since $\frac{x}{2\sin(x)\cos(x)}\sim 1/2$, but $\lim_{x\to 0}\cos(1/x)$ fails to exist as $x\to 0$.
The derivative of the numerator is $$f'(x)=3x^2\sin\frac{1}{x}-x\cos\frac{1}{x}$$ which has indeed limit $0$ at $0$; also the derivative of the denominator, which is $g'(x)=\sin2x$ has limit $0$. However, another application fails: $$f''(x)=6x\sin\frac{1}{x}-4\cos\frac{1}{x}+\frac{1}{x}\sin\frac{1}{x}$$ which has no limit at $0$.
However the original limit exists and is $0$: $$\lim_{x\to0}\frac{x^2}{\sin^2x}x\sin\frac{1}{x}=1\cdot0=0$$
• @Dr.MV The limit of the denominator exists (and isn't $0$), so the fact that the numerator has no limit is sufficient. – egreg Nov 9 '16 at 16:35