2. Find the real numbers $a, b, c>0$ if

$$
\lim _{n \rightarrow \infty}\left(\sqrt{a^{2} n^{2}+2014 n+1}-b n+c\right)=\sqrt{\frac{2 c}{a}} \cdot \sqrt{2014} \text { and } a+c=72 \text {. }
$$

Prof.Voiculeț Septimius, Videle
2. Case 1) $ab \Rightarrow \lim _{n \rightarrow \infty} n\left(\sqrt{a^{2}+\frac{2014}{n}+\frac{1}{n^{2}}}-b+\frac{c}{n}\right)=\infty(a-b)=\infty$ does not work..........1p

Case 3) $a=b$

$\Rightarrow \lim _{n \rightarrow \infty} \frac{a^{2} n^{2}+2014 n+1-a^{2} n^{2}+2 a c n-c^{2}}{\sqrt{a^{2} n^{2}+2014 n+1}+a n-c}=\lim _{n \rightarrow \infty} \frac{n\left(2014+2 a c+\frac{1-c^{2}}{n}\right)}{n\left(\sqrt{a^{2}+\frac{2014}{n}+\frac{1}{n^{2}}}+a-\frac{c}{n}\right)}=\frac{2014+2 a c}{2 a}$ $2 p$

$\frac{2014+2 a c}{2 a}=\sqrt{\frac{2 c}{a}} \cdot \sqrt{2014} \Leftrightarrow \frac{1007}{a}+c=2 \sqrt{\frac{1007}{a}} \cdot c \Leftrightarrow \frac{1007}{a}=c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

$a c=1007$ and $a+c=72 \Leftrightarrow(a=53, c=19)$ or $(a=19, c=53) \ldots \ldots \ldots .1 \mathrm{~m}$

This results in two solutions $(a=53, b=53, c=19),(a=19, b=19, c=53) \ldots \ldots . .1 p$