Prove if sequence is convergent then $a = 1$

Question

Let $a, b, c \in \mathbb{R}$ be such that $a > 0$ and $b^2 < 4ac$. Consider the sequence:

$x_n = \sqrt{an^2 + bn +c} -n$ , $n\ge1$

I'm supposed to prove that if $x_n$ is convergent then $a = 1$. My hint is that the conjugate will help but I'm not really sure where to start. Any help is appreciated.

Answer 1

With conjugate write $$\lim_\infty\sqrt{an^2+bn+c}-n=\lim_\infty\dfrac{(a-1)n^2+bn+c}{\sqrt{an^2+bn+c}+n}=\lim_\infty\dfrac{(a-1)n^2+bn+c}{n\left(\sqrt{a+\frac{b}{n}+\frac{c}{n^2}}+1\right)}$$ exists if $a-1=0$.

Answer 2

I would just pull a factor $n\sqrt a$ out of the square root, then show the remaining square root goes to $1$. The expression then goes as $n(\sqrt a-1)$ which diverges unless $a=1$