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?
 A: Here is a slightly different approach.
First note that 
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b}{x^1}&=\lim_{x\to \infty}\sqrt {1-\frac6x+\frac3{x^2}}-a+\frac bx\\
&=1-a
\end{align}
and 
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b-(1-a)x}{x^0}&=\lim_{x\to \infty}\sqrt {x^{2}-6x+3}+b-x\\
&=\lim_{x\to \infty}\frac{2(b-3)x+3-b^2}{\sqrt {x^{2}-6x+3}-b+x}\\
&=b-3
\end{align}
and finally 
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b-(1-a)x-(b-3)}{x^{-1}}&=...\\
&=-3
\end{align}
Putting all this together we see that $$\sqrt {x^{2}-6x+3}-ax+b=(1-a)x+(b-3)-\frac3x+O(\frac1{x^2}).$$
The conclusion is yours!
A: Note that
$$
\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right|
$$
is meaningless, because the limit cannot depend on $x$.
One way can be doing a substitution, $x=-1/t$, so the limit becomes
$$
\lim_{t\to0^+}\sqrt{\frac{1}{t^2}-\frac{6}{t}+3}-\frac{a}{t}+b=
\lim_{t\to0^+}\frac{\sqrt{1-6t+3t^2}-a+bt}{t}
$$
In order that the limit is finite, we must have that the numerator has limit $0$; since the numerator is continuous, this means $a=1$. Now the limit is just the derivative at $0$ of the function
$$
t\mapsto \sqrt{1-6t+3t^2}-1+bt
$$
so compute at $0$ the expression
$$
\frac{-6+6t}{2\sqrt{1-6t+3t^2}}+b
$$
which gives
$$
-3+b=1
$$
Thus $a=1$ and $b=4$.
A: 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 $a = 1, b = 4$
