\begin{gather*} \lim_{x\to\infty}\dfrac{x^{2}-x}{\sqrt{x^{2}+x}}=\lim\sqrt{\dfrac{(x^{2}-x)^{2}}{x^{2}+x}}=\lim\sqrt{\dfrac{x^{4}-2x^{3}+x^{2}}{x^{2}+x}}=\\ =\lim\sqrt{\dfrac{\tfrac{x^{4}}{x^{4}}-2\tfrac{x^{3}}{x^{4}}+\tfrac{x^{2}}{x^{4}}}{\tfrac{x^{2}}{x^{4}}+\tfrac{x}{x^{4}}}}=\lim\sqrt{\dfrac{1-\tfrac{2}{x}+\tfrac{1}{x^{2}}}{\tfrac{1}{x^{2}}+\tfrac{1}{x^{3}}}}=\lim\sqrt{\dfrac{1-0+0}{0+0}}=\lim\sqrt{\dfrac{1}{0}}=\\ =\lim\dfrac{1}{+0}=+\infty\end{gather*} Megjegyzes: A $\lim\limits_{x\to\infty}\dfrac{1}{x^{k}}=0$, ha $k\in\mathbb{Z}^{+}$.