Neodređeni oblik

S obzirom da je

$\displaystyle \lim_{x\to1\pm 0}\frac{1}{1-x}=\mp\infty \quad\textrm{i}\quad \lim_{x\to1\pm0}\frac{3}{1-x^3}=\mp\infty,$

računanje limesa zdesna i slijeva u

su daje neodređeni oblik $\infty -\infty$ . Svođenjem na zajednički nazivnik dobivamo

$\displaystyle \lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$	$\displaystyle = \lim_{x\to1}\frac{x^2+x-2}{1-x^3}$
	$\displaystyle = \lim_{x\to1}\frac{(x-1)\cdot(x+2)}{-(x-1)\cdot(x^2+x+1)}$
	$\displaystyle = \lim_{x\to1}\frac{x+2}{-(x^2+x+1)}=-1.$

Očigledno je

$\displaystyle \lim_{x\to-\infty}\left(x-\sqrt{x^2+1}\right)=-\infty-\infty=-\infty.$

S druge strane, računanje $\displaystyle\lim_{x\to+\infty}\left(x-\sqrt{x^2+1}\right)$ daje neodređeni oblik $\infty -\infty$ . Racionalizacijom dobivamo

$\displaystyle \lim_{x\to+\infty}\left(x-\sqrt{x^2+1}\right)$	$\displaystyle =\lim_{x\to+\infty}\left(x-\sqrt{x^2+1}\right)\cdot\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$
	$\displaystyle = \lim_{x\to+\infty}\frac {x^2-(x^2+1)}{x+\sqrt{x^2+1}}$
	$\displaystyle = \lim_{x\to+\infty}\frac {-1}{x+\sqrt{x^2+1}}=0.$

S obzirom da je

$\displaystyle \lim_{x\to\pm\infty}\sqrt{x^2-2x-1}=+\infty\quad\textrm{i}\quad \lim_{x\to\pm\infty}\sqrt{x^2-7x+3}=+\infty,$

računanje zadanih limesa daje neodređeni oblik $\infty -\infty$ . Racionalizacijom dobivamo

	$\displaystyle \quad\,\lim_{x\to\pm\infty}\left(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3}\right)$
	$\displaystyle = \lim_{x\to \pm \infty}\left(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3}\rig... ...e\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}{\displaystyle\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}$
	$\displaystyle = \lim_{x\to \pm \infty}\frac{(x^2-2x-1)-(x^2-7x+3)}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}$
	$\displaystyle = \lim_{x\to \pm \infty}\frac{5x-4}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}$
	$\displaystyle = \lim_{x\to\pm\infty}\frac{ x\left(5-\frac{4}{x}\right)}{ \vert ... ...t(\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}+\sqrt{1-\frac{7}{x}+\frac{3}{x^2}}\right)}$
	$\displaystyle = \lim_{x\to\pm\infty}\frac{x}{\vert x\vert}\cdot \lim_{x\to\pm\i... ...{x}} { \sqrt{1-\frac{2}{x}+\frac{1}{x^2}}+ \sqrt{1-\frac{7}{x}+\frac{3}{x^2}}}.$

Budući je

$\displaystyle \lim_{x\to +\infty}\frac{x}{\vert x\vert}=1 \quad\textrm{i}\quad \lim_{x\to -\infty}\frac{x}{\vert x\vert}=-1,$

iz dobivenog rezultata slijedi

$\displaystyle \lim_{x\to \pm \infty} \left(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3}\right)= \begin{cases}\;\;\, 5/2, & x\to +\infty \\ -5/2, & x\to -\infty. \end{cases}$