Limit $\lim \limits_{x\to -\infty }\frac{\sqrt{16x^{2}+2x-3} }{x+9}$ Given $$\lim \limits_{x\to -\infty }\frac{\sqrt{16x^{2}+2x-3} }{x+9} $$
Hi,  I need help for proof this limit, which could be used arguments or results. I would appreciate any suggestions. I Don't use l'Hopital rule. 
 A: $$\lim _{ x\to -\infty  } \frac { \sqrt { 16x^{ 2 }+2x-3 }  }{ x+9 } =\lim _{ x\to -\infty  } \frac { \left| x \right| \sqrt { 16+\frac { 2 }{ x } -\frac { 3 }{ { x }^{ 2 } }  }  }{ x\left( 1+\frac { 9 }{ x }  \right)  } =\lim _{ x\to -\infty  } \frac { -x\sqrt { 16+\frac { 2 }{ x } -\frac { 3 }{ { x }^{ 2 } }  }  }{ x\left( 1+\frac { 9 }{ x }  \right)  } =-4$$
A: As some have mentioned in the comments to your question. The key strategy is to divide numerator and denominator by $x$:
$$
\lim_{x\to -\infty}\frac{\sqrt{16x^{2}+2x-3}}{x+9} = \lim_{x\to -\infty}\frac{\sqrt{x^2(16+2/x-3/x^2)}}{x+9} = \lim_{x\to -\infty}\frac{\lvert x\rvert\sqrt{16+2/x-3/x^2}}{x+9} \\
= -\lim_{x\to -\infty}\frac{\sqrt{16+2/x-3/x^2}}{1+9/x}.
$$
Can you see what happens in the limit as $x \to -\infty$?
A: I suggest first simplify the denominator by changing of variable $t = x + 9$, then the numerator becomes
$$\sqrt{16(t - 9)^2 + 2(t - 9) - 3} = \sqrt{16t^2 - 286t + 1275}.$$
Hence the limit to be evaluated is 
\begin{align*}
\lim_{t \to -\infty} \frac{\sqrt{16t^2 - 286t + 1275}}{t} = 
-\lim_{t \to -\infty} \sqrt{16 - \frac{286}{t} + \frac{1275}{t^2}} = -4.
\end{align*}
