Odpowiedź:
[tex]\huge\boxed{1)\ \log8}\\\boxed{2)\ \dfrac{1}{2}}[/tex]
Szczegółowe wyjaśnienie:
[tex]1)\\\lim\limits_{n\to\infty}\bigg[4\log(2n)-\log\dfrac{2n^2-1}{3}-\log(3n^2-1)\bigg][/tex]
skorzystamy z twierdzeń:
[tex]\log_ab^n=n\log_ab\\\\\log_ab-\log_ac=\log_a\dfrac{b}{c}[/tex]
[tex]=\lim\limits_{n\to\infty}\bigg[\log\bigg((2n)^4:\dfrac{2n^2-1}{3}:(3n^2-1)\bigg)\bigg]\\\\=\lim\limits_{n\to\infty}\bigg[\log\bigg(16n^4\cdot\dfrac{3}{2n^2-1}\cdot\dfrac{1}{3n^2-1}\bigg)\bigg]\\\\=\lim\limits_{n\to\infty}\bigg(\log\dfrac{48n^4}{6n^4-5n^2+1}\bigg)\\\\=\lim\limits_{n\to\infty}\bigg(\log\dfrac{48n^4}{n^4\left(6-\frac{5}{n^2}+\frac{1}{n^4}\right)}\bigg)\\\\=\lim\limits_{n\to\infty}\bigg(\log\dfrac{48}{6-\frac{5}{n^2}+\frac{1}{n^4}}\bigg)=\log\dfrac{48}{6}=\log8[/tex]
[tex]2)\\\lim\limits_{n\to\infty}\dfrac{2^{\sqrt{n^2+n}}}{2^{\sqrt{n^2+3n}}}=\lim\limits_{n\to\infty}2^{\sqrt{n^2+n}-\sqrt{n^2+3n}}=2^{\lim\limits_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+3n})}=(*)[/tex]
Skorzystamy ze wzoru:
[tex]\dfrac{a^n}{a^m}=a^{n-m}[/tex]
Zajmijmy się na razie tylko granicą wykładnika:
[tex]\lim\limits_{n\to\infty}\left(\sqrt{n^2+n}-\sqrt{n^2+3n}\right)\\\\=\lim\limits_{n\to\infty}\dfrac{(\sqrt{n^2+n}-\sqrt{n^2+3n})(\sqrt{n^2+n}+\sqrt{n^2+3n})}{\sqrt{n^2+n}+\sqrt{n^2+3n}}\\\\=\lim\limits_{n\to\infty}\dfrac{(\sqrt{n^2+n})^2-(\sqrt{n^2+3n})^2}{\sqrt{n^2+n}+\sqrt{n^2+3n}}\\\\=\lim\limits_{n\to\infty}\dfrac{n^2+n-n^2-3n}{\sqrt{n^2+n}+\sqrt{n^2+3n}}[/tex]
[tex]=\lim\limits_{n\to\infty}\dfrac{-2n}{\sqrt{n^2(1+\frac{1}{n})}+\sqrt{n^2(1+\frac{3}{n})}}\\\\=\lim\limits_{n\to\infty}\dfrac{-2n}{n\sqrt{1+\frac{1}{n}}+n\sqrt{1+\frac{3}{n}}}\\\\=\lim\limits_{n\to\infty}\dfrac{-2n}{n\left(\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{3}{n}}\right)}\\\\=\lim\limits_{n\to\infty}\dfrac{-2}{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{3}{n}}}=\dfrac{-2}{\sqrt1+\sqrt1}=\dfrac{-2}{1+1}=\dfrac{-2}{2}=-1[/tex]
Podstawiamy do [tex](*)[/tex]:
[tex](*)=2^{-1}=\dfrac{1}{2}[/tex]
