\[ \begin{align} \lim_{x\rightarrow \infty}\;{\left(\frac{x^{3}}{2x^{2}-1}-\frac{x^{2}}{2x+1}\right)} &=\lim_{x\rightarrow \infty}\;{\left[\frac{x^{3}(2x+1)-x^{2}(2x^{2}-1)}{(2x^{2}-1)(2x+1)}\right]}=\\[5pt] &=\lim_{x\rightarrow \infty}\;{\left[\frac{x^{3}(2x+1)-x^{2}(2x^{2}-1)}{(2x^{2}-1)(2x+1)}\right]}=\\[5pt] &=\lim_{x\rightarrow \infty}\;{\left[\frac{2x^{4}+x^{3}-2x^{4}+x^{2}}{4x^{3}+2x^{2}-2x-1}\right]}=\\[5pt] &=\lim_{x\rightarrow \infty}\;{\left[\frac{x^{3}+x^{2}}{4x^{3}+2x^{2}-2x-1}\right]}=\\[5pt] &=\lim_{x\rightarrow \infty}\;{\left[\frac{\cancel{x^{3}}\left(1+\dfrac{1}{x}\right)}{\cancel{x^{3}}\left(4+\dfrac{2}{x}-\dfrac{2}{x^{2}}-\dfrac{1}{x^{3}}\right)}\right]} \end{align} \]

\[ \lim_{x\rightarrow \infty}\;{\left(\frac{x^{3}}{2x^{2}-1}-\frac{x^{2}}{2x+1}\right)}=\lim_{x\rightarrow \infty }\;{\left(\frac{1+\cancelto{0}{\frac{1}{\infty}}}{4+\cancelto{0}{\frac{2}{\infty}}-\cancelto{0}{\frac{2}{\infty ^{2}}}-\cancelto{0}{\frac{1}{\infty^{3}}}}\right)}=\frac{1}{4} \]