Without using l'hopital's rule 2

$\begin{aligned} L & = \lim_{x \to 0} \frac {x^2}{\sin x} & \small \color{#3D99F6} \text{Divide up and down by }x \\ & = \lim_{x \to 0} \frac {x}{\color{#3D99F6} \frac {\sin x}x} & \small \color{#3D99F6} \text{See note.} \\ & = \frac 0{\color{#3D99F6}1} = \boxed{0} \end{aligned}$

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Note:

$\begin{aligned} \lim_{x \to 0} \frac {\color{#3D99F6}\sin x}x & = \lim_{x \to 0} \frac {\color{#3D99F6}x-\frac {x^3}{3!}+\frac{x^5}{5!}-...}x & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \left(1-\frac {x^2}{3!}+\frac{x^4}{5!}-... \right) \\ & = \boxed{1} \end{aligned}$

Let $x=0.00001$ , then we have

$\dfrac{0.00001^2}{\sin 0.00001}=0$

Note: Calculator must be in radian mode.