# 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.

• Divide numerator and denominator by $x$ and see if you can simplify and then evaluate the limit. – rogerl Aug 24 '16 at 21:30
• HINT: The equivalent of the square root term is $4|x|$ while the equivalent of $x+9$ is $x$. – Mark Viola Aug 24 '16 at 21:31

$$\lim _{ x\to -\infty } \frac { \sqrt { 16x^{ 2 }+2x-3 } }{ x+9 } =\lim _{ x\to -\infty } \frac { \left| x \right| \sqrt { 16+\frac { 2 }{ x } -\frac { 3 }{ { x }^{ 2 } } } }{ x\left( 1+\frac { 9 }{ x } \right) } =\lim _{ x\to -\infty } \frac { -x\sqrt { 16+\frac { 2 }{ x } -\frac { 3 }{ { x }^{ 2 } } } }{ x\left( 1+\frac { 9 }{ x } \right) } =-4$$

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 $x \to -\infty$?

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*}