Rozwiązanie

$\underset{n\to \infty }{\lim }(n+5{)}^2\left(\frac{p+1}{(n+1)(n+2)}+\frac{2p+2}{(n+2)(n+3)}\right)=$

$=\underset{n\to \infty }{\lim }\left((p+1)\cdot \frac{(n+5{)}^2}{(n+1)(n+2)}+(2p+2)\cdot \frac{(n+5{)}^2}{(n+2)(n+3)}\right)=$

$=\underset{n\to \infty }{\lim }\left((p+1)\cdot \frac{n^2+10n+25}{n^2+3n+2}+(2p+2)\cdot \frac{n^2+10n+25}{n^2+5n+6}\right)=$

$=\underset{n\to \infty }{\lim }\left((p+1)\cdot \frac{1+\frac{10}{n}+\frac{25}{n^2}}{1+\frac{3}{n}+\frac{2}{n^2}}+(2p+2)\cdot \frac{1+\frac{10}{n}+\frac{25}{n^2}}{1+\frac{5}{n}+\frac{6}{n^2}}\right)=(p+1)\cdot 1+(2p+2)\cdot 1=3p+3$

$3p+3=12$    $\to$    $p=3$

Odp.: Granica jest równa $12$ dla $p=3$.

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