Rozwiązanie

$\underset{n\to \infty }{\lim }\left(\frac{n^2}{n+2}-\frac{(n+2{)}^2}{n+444}\right)=[\infty -\infty ]$    $\leftarrow$    symbol nieoznaczony

$\underset{n\to \infty }{\lim }\left(\frac{n^2}{n+2}-\frac{(n+2{)}^2}{n+444}\right)=\underset{n\to \infty }{\lim }\frac{n^2(n+444)-(n+2{)}^3}{(n+2)(n+444)}=$

$=\underset{n\to \infty }{\lim }\frac{n^3+444n^2-(n^3+6n^2+12n+8)}{n^2+446n+888}=\underset{n\to \infty }{\lim }\frac{438n^2-12n-8}{n^2+446n+888}=$

$=\underset{n\to \infty }{\lim }\frac{438-\frac{12}{n}-\frac{8}{n^2}}{1+\frac{446}{n}+\frac{888}{n^2}}=\frac{438}{1}=438$

Odp.: Granicą jest $438$.