## Problem 1

Determine $a, b, c \in \mathbb{R}$, for which $\lim _{n \rightarrow \infty} n\left(\sqrt{a n^{2}+b n+c}-2 n-3\right)=2015$.
## Problem 1

We denote the limit by $l$.

$$
\begin{aligned}
& l=\lim _{n \rightarrow \infty} n^{2}\left[\sqrt{a+\frac{b}{n}+\frac{c}{n^{2}}}-\left(2+\frac{3}{n}\right)\right] \Rightarrow \text{for} \sqrt{a}-2 \neq 0 \Rightarrow l= \pm \infty \neq 2015, \text { does not fit } \Rightarrow \sqrt{a}=2 \Rightarrow a=4 . .2 \mathbf{p} \\
& l=\lim _{n \rightarrow \infty} n \cdot \frac{4 n^{2}+b n+c-4 n^{2}-12 n-9}{\sqrt{4 n^{2}+b n+c}+2 n+3}=\lim _{n \rightarrow \infty} \frac{(b-12) n^{2}+(c-9) n}{n\left(\sqrt{4+\frac{b}{n}+\frac{c}{n^{2}}}+2+\frac{3}{n}\right)} \text {, for } b-12 \neq 0 \Rightarrow l= \pm \infty \neq 2015 \\
& \text { does not fit } \Rightarrow b=12 \\
& 3 \mathbf{2} \\
& l=\lim _{n \rightarrow \infty} \frac{(c-9) \not n}{\not n\left(\sqrt{4+\frac{b}{n}+\frac{c}{n^{2}}}+2+\frac{3}{n}\right)}=\frac{c-9}{4} \Rightarrow \frac{c-9}{4}=2015 \Rightarrow c=8069
\end{aligned}
$$