If $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $ then find $a+b$ $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $
if I use this $$\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right| $$
I find $a=1,b=4$ but if I try to multiple by its conjugate
$$\lim _{x\rightarrow -\infty }\dfrac {x^{2}+6x+3-\left( ax+b\right) ^{2}} {\left| x\right| \sqrt {1+\dfrac {6} {x}+\dfrac {3} {x^{2}}}-ax-b}=\lim _{x\rightarrow -\infty } \dfrac {x\left( x+6+\dfrac {3} {x}-a^{2}x-2ab-\dfrac {b^{2}} {x}\right) } {x\left( -\sqrt {1+\dfrac {6} {x}+\dfrac { {3}} {x^{2}}}-a-\dfrac {b} {x}\right) }$$
in order to lim of this equal to $1$ $$x\left(\underbrace { 1-a^{2}}_0\right) +6-2ab=-1-a$$
if $a=1$ then $ b=4$
if $a=-1$ so $b=-3$ but wolfram says $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}-x-3=\infty$
can u tell me where is my mistake?

Let's start this step by step, $\lim_{x\to -\infty} \sqrt{x^2 + bx + c} + x= \lim_{x\to -\infty} \sqrt{(x + \frac{b}{2})^2 + c - \frac{b^2}{4}} + x = \lim_{x\to -\infty} |x+\frac{b}{2}| +x = -\frac{b}{2}$ Now everything is simpler than ever, $\lim_{x\to -\infty} \sqrt(x^2 + 6x + 3) + x +(a-1)x b = \lim_{x\to -\infty} -3+b + (a-1)x = 1$, now it is clear that $a = 1, b = 4$
