 Uprav�me v�raz $(x^2-1)/(2x^2-x-1)$:
   $$
      \frac{x^2-1}{2x^2-x-1} = \frac{(x-1)(x+1)}{(x-1)(2x+1)} =
      \frac{x+1}{2x+1}
   $$
   Te� m��eme spo��st limity:
   \begin{eqnarray*}
      \lim_{x\to0}\frac{x+1}{2x+1} & = & 1         \\
      \lim_{x\to1}\frac{x+1}{2x+1} & = & \frac23   \\
      \lim_{x\to\infty}\frac{x+1}{2x+1} & = &
      \lim_{x\to0_+}\frac{\frac1x+1}{\frac2x+1} \; = \;
      \lim_{x\to0_+}\frac{\frac{x+1}x}{\frac{2+x}x} \; = \;
      \lim_{x\to0_+}\frac{x+1}{x+2} \; = \; 0.
   \end{eqnarray*}