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Solving without L'Hopital $$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$

That's

$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}\right)-\lim_{x\to-\infty}\left(\sqrt{4x^2+x}\right)$$

I have been taught to get the highest exponents, so...

$$\sqrt{4x^2}-\sqrt{4x^2}$$

It's the same for both sides, no?

$$2x-2x$$

$$-\infty+\infty$$

Which is wrong. The correct answer is

$$\frac{1}{4}$$

Why? I always just grab the highest exponent ($\sqrt{4x^2}$) and work with it. But this time it didn't go well.

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  • $\begingroup$ Already your first step is wrong. The rule $\lim(f+g)=\lim f + \lim g$ is only valid if the two limits on the right-hand side are finite. $\endgroup$ – Hans Lundmark Sep 28 '15 at 7:14
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Notice, $$\lim_{x\to -\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$ $$=\lim_{x\to \infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)$$ $$=\lim_{x\to \infty}\frac{\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$ $$=\lim_{x\to \infty}\frac{4x^2-6-4x^2+x}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$ $$=\lim_{x\to \infty}\frac{x-6}{x\left(\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}\right)}$$ $$=\lim_{x\to \infty}\frac{1-\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}}$$ $$=\frac{1-0}{\sqrt{4-0}+\sqrt{4-0}}=\frac{1}{2+2}=\color{red}{\frac{1}{4}}$$

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  • $\begingroup$ $\dfrac{1}{2+2} = \dfrac{1}{4}$ $\endgroup$ – DeepSea Sep 28 '15 at 4:51
  • $\begingroup$ After rationalizing, shouldn't it have been $4x^2-6-4x^2\color{red}{-x}$ instead of $4x^2-6-4x^2\color{red}{+x}$? $\endgroup$ – Zol Tun Kul Sep 28 '15 at 5:51
  • $\begingroup$ Notice, $(\sqrt{4x^2-6}-\sqrt{4x^2\color{red}{-x}})(\sqrt{4x^2-6}+\sqrt{4x^2\color{red}{-x}})=4x^2-6-(4x^2\color{red}{-x})$$$=4x^2-6-4x^2\color{red}{+x}$$ $\endgroup$ – Harish Chandra Rajpoot Sep 28 '15 at 6:30
  • $\begingroup$ But isn't it $\sqrt{4x^2+x}$ instead of $\sqrt{4x^2-x}$? $\endgroup$ – Zol Tun Kul Sep 29 '15 at 4:49
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    $\begingroup$ No, because we substitute $x=-x$ then $x\to +\infty$ $\endgroup$ – Harish Chandra Rajpoot Sep 29 '15 at 5:11
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hint: Use the identity: $\sqrt{A}-\sqrt{B} = \dfrac{A-B}{\sqrt{A}+\sqrt{B}}$, and $\sqrt{4x^2\pm6} = x\sqrt{4\pm\dfrac{6}{x^2}}$

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  • $\begingroup$ Does the first identity have a name I can search for? $\endgroup$ – Zol Tun Kul Sep 28 '15 at 4:52
  • $\begingroup$ Its a variant of the well known $A^2-B^2 = (A-B)(A+B)$. $\endgroup$ – DeepSea Sep 28 '15 at 5:09
  • $\begingroup$ I used the two things you showed me and progressed to $$\frac{-6-x}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}$$. While it does seem a bit better it is still not quite clear to me what to do with that. $\endgroup$ – Zol Tun Kul Sep 28 '15 at 5:18
  • $\begingroup$ You missed the $x$ in front of each square root, and divide the top by $x$. $\endgroup$ – DeepSea Sep 28 '15 at 5:26
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\begin{align} \lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)&=\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)\cdot\frac{\sqrt{4x^2-6}+\sqrt{4x^2+x}}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\ &=\lim_{x\to-\infty}\frac{(\sqrt{4x^2-6})^2-(\sqrt{4x^2+x})^2}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\ &=\lim_{x\to-\infty}\frac{4x^2-6-(4x^2+x)}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\ &=\lim_{x\to-\infty}\frac{(-x-6)/|x|}{\sqrt{4-6/x^2}+\sqrt{4+x/x^2}}\\ &=\lim_{x\to-\infty}\frac{1+\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}\\ &=\frac{1}{\sqrt{4}+\sqrt{4}}\\ &=\frac{1}{4} \end{align}

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  • $\begingroup$ When you divide $(-x-6)/|x|$, why did you pick $|x|$ instead of $x$? Why is this allowed? Also, how can $(-x-6)/|x|$ become $1+\frac{6}{x}$? Dividing by infinity should be $0$, I think. $\endgroup$ – Zol Tun Kul Sep 29 '15 at 5:06
  • $\begingroup$ @ZolTunKul: In order to make manipulations inside the surds. $\endgroup$ – Ángel Mario Gallegos Sep 29 '15 at 14:16
  • $\begingroup$ I'm having trouble seeing how can $(-x-6) / |x|$ result in $(1 + 6/x)$. Since $|x| = +\infty$, which is positive, how come the signs are shifted? I mean, $-x$ becomes $1$ (negative to positive) and $-6$ becomes $6/x$ (negative to positive). Shouldn't $(-x-6)$ be divided by just $x$? Since $x = -\infty$ then the signs can indeed by shifted. I think. $\endgroup$ – Zol Tun Kul Sep 30 '15 at 15:44
  • $\begingroup$ @ZolTunKul: Notice $x\to -\infty$ is not the same that $x=-\infty$ as it seems you are supposing. Instead, as $x\to -\infty$ we have that we can make $|x|$ as large as we want and $x<0$. So, $|x|=-x$, since $x<0$, then $$\frac{-x-6}{|x|}=\frac{-x-6}{-x}=\frac{x+6}{x}=1+\frac{6}{x}$$ $\endgroup$ – Ángel Mario Gallegos Sep 30 '15 at 15:52
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Set $-1/x=y\implies y\to0^+, |y|=+y$

and $4x^2-6=\dfrac{4-6y^2}{y^2},\sqrt{4x^2-6}=\dfrac{\sqrt{4-6y^2}}{|y|}=\dfrac{\sqrt{4-6y^2}}y$

$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)=\lim_{y\to0^+}\dfrac{\sqrt{4-6y^2}-\sqrt{4-y}}y$$

$$=\lim_{y\to0^+}\dfrac{(4-6y^2)-(4-y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$

$$=\lim_{y\to0^+}\dfrac{y(1-6y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$

Cancel out $y$ as $y\ne0$ as $y\to0$

Then set $y=0$

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