1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Divide numerator and denominator by $x$ and see if you can simplify and then evaluate the limit. $\endgroup$ – rogerl Aug 24 '16 at 21:30
  • 2
    $\begingroup$ HINT: The equivalent of the square root term is $4|x|$ while the equivalent of $x+9$ is $x$. $\endgroup$ – Mark Viola Aug 24 '16 at 21:31
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $x \to -\infty$?

$\endgroup$
0
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.