$Let \underset{x\to 0}{\lim }\frac{{\left[\mathrm{x}\right]}^2}{{\mathrm{x}}^2}=\mathrm{l} \mathrm{and} \mathrm{Let} \underset{\mathrm{x}\to 0}{\lim }\frac{{\displaystyle {\left[\mathrm{x}\right]}^2}}{{\displaystyle {\mathrm{x}}^2}}=\mathrm{m}$, where [.] denotes greatest integer. Then,

$\begin{array}{l}\frac{{\left[\mathrm{x}\right]}^2}{{\mathrm{x}}^2}=\left[\begin{array}{l}0, \mathrm{if} 0<\mathrm{x}<1\\ \frac{1}{{\mathrm{x}}^2}, \mathrm{if} -1<\mathrm{x}<0\end{array}\right.\Rightarrow \mathrm{l} \mathrm{does} \mathrm{not} \mathrm{exist}\\ \frac{\left[{\mathrm{x}}^2\right]}{{\mathrm{x}}^2}=\left[\begin{array}{l}0, \mathrm{if} 0<\mathrm{x}<1\\ 0, \mathrm{if} -1<\mathrm{x}<0\end{array}\right.\Rightarrow \mathrm{m} \mathrm{exists} \mathrm{and} \mathrm{is} \mathrm{equal} \mathrm{to} 0\end{array}$