Granicę ciągu nieskończonego liczymy przy n→∞
[tex]\bold{\lim\limits_{n\to\infty}\dfrac{4^n-1}{2^n+6}= \lim\limits_{n\to\infty}\dfrac{\,\frac{4^n}{4^n}-\frac1{4^n}\,}{\frac{2^n}{4^n}+\frac6{4^n} } =\lim\limits_{n\to\infty}\dfrac{\,1-\left\{\frac1{4^n}\right\}^{\nearrow0}\,}{\left\{\left(\frac12\right)^n\right\}_{\searrow0}+\left\{\frac6{4^n}\right\}_{\searrow0} } =\left[\dfrac10\right]=\infty}[/tex]
[tex]\bold{\lim\limits_{n\to\infty}\dfrac{4-6^n}{4^n-6}= \lim\limits_{n\to\infty}\dfrac{\,\frac{4}{6^n}-\frac{6^n}{6^n}\,}{\frac{4^n}{6^n}-\frac6{6^n} } =\lim\limits_{n\to\infty}\dfrac{\,\left\{\frac4{6^n}\right\}^{\nearrow0}-1\,}{\left\{\left(\frac23\right)^n\right\}_{\searrow0}-\left\{\frac6{6^n}\right\}_{\searrow0} } =\left[\dfrac{-1}0\right]=-\infty}[/tex]
[tex]\bold{\lim\limits_{n\to\infty}\dfrac{4^n-3\cdot8^n+2}{2^n+2}= \lim\limits_{n\to\infty}\dfrac{\,\frac{4^n}{8^n}-3\cdot\frac{8^n}{8^n}+\frac2{8^n}\,}{\frac{2^n}{8^n}+\frac2{8^n} } =}\\\\\\\bold{=\lim\limits_{n\to\infty}\dfrac{\,\left\{\left(\frac12\right)^n\right\}^{\nearrow0}-3\cdot1+\left\{\frac2{8^n}\right\}^{\nearrow0}\,}{\left\{\left(\frac14\right)^n\right\}_{\searrow0}+\left\{\frac2{8^n}\right\}_{\searrow0} } =\left[\dfrac{-3}0\right]=-\infty}[/tex]
