Can l'Hôpital's rule be applied to L = lim as x to 0

Can l'Hôpital's rule be applied to $\displaystyle L = \lim_{x\to0}\frac{x^{2}\sin\dfrac1x}{\sin x}$? Find the limit.

Step 1: The form is right

Since $\left|x^{2}\sin\dfrac1x\right|\le x^{2}\to0$ and $\sin x\to0$, the quotient is of the form $\dfrac00$, so the rule may in principle be applied.

Step 2: Attempt the rule

Differentiating the numerator by the product and chain rules: \[\frac{d}{dx}\left(x^{2}\sin\frac1x\right) = 2x\sin\frac1x+x^{2}\cos\frac1x\cdot\left(-\frac{1}{x^{2}}\right) = 2x\sin\frac1x-\cos\frac1x,\]

while $\dfrac{d}{dx}\sin x = \cos x$. The rule would require

\[ \lim_{x\to0}\frac{2x\sin\dfrac1x-\cos\dfrac1x}{\cos x}. \tag{1} \]

Here $\cos x\to1$ and $2x\sin\dfrac1x\to0$, but $\cos\dfrac1x$ oscillates between $-1$ and $1$ ever faster as $x\to0$ and has no limit. Hence (1) does not exist and the rule cannot be applied.

Step 3: Evaluate by elementary means

Factor the expression so that each piece is recognisable: \[\frac{x^{2}\sin\dfrac1x}{\sin x} = x\cdot\frac{x}{\sin x}\cdot\sin\frac1x. \tag{2}\]

As $x\to0$: the first factor tends to $0$, the second to $1$ (a standard limit), and the third is bounded by $1$ in absolute value. A quantity tending to zero times a bounded quantity tends to zero:

\[ \left|\frac{x^{2}\sin\dfrac1x}{\sin x}\right| \le |x|\cdot\frac{|x|}{|\sin x|} \longrightarrow 0. \tag{3} \]

Step 4: Results

\[\text{L'H\^opital's rule is inapplicable: } \cos\frac1x \text{ oscillates without limit.}\]

Step 5: Verification

\[x = 0.1:\ -0.0544929; \qquad x = 0.01:\ -0.0050637; \qquad x = 0.001:\ 0.0008269 \;\;\checkmark\]

\[\text{The values change sign erratically but obey } \left|\,\cdot\,\right|\le\frac{x^{2}}{|\sin x|}\approx|x| \;\;\checkmark\]

\[\text{Even function: } f(-x) = f(x),\ \text{and } x = -0.01 \text{ gives } -0.0050637 \text{ as well} \;\;\checkmark\]