Oblik $a^\infty$

Izračunajte

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

Rješenje. Budući je

$\displaystyle \lim_{x\to\infty}\frac{x^2+2}{2x^2+1}=\lim_{x\to\infty}\frac{1+\d... ...yle \frac{1}{x^2}}=\frac{1}{2}\quad\textrm{i}\quad \lim_{x\to\infty}x^2=\infty,$

vrijedi

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

jer je

$\displaystyle \lim_{t\to+\infty}a^t= \left\{ \begin{matrix} 0, & 0<a<1\\ +\infty, & a>1 \end{matrix} \right..$

Još vrijedi

$\displaystyle \lim_{t\to-\infty}a^t= \left\{ \begin{matrix} +\infty, & 0<a<1\\ 0, & a>1 \end{matrix} \right..$