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

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.

• 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. – Hans Lundmark Sep 28 '15 at 7:14

Notice, $$\lim_{x\to -\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$ $$=\lim_{x\to \infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)$$ $$=\lim_{x\to \infty}\frac{\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$ $$=\lim_{x\to \infty}\frac{4x^2-6-4x^2+x}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$ $$=\lim_{x\to \infty}\frac{x-6}{x\left(\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}\right)}$$ $$=\lim_{x\to \infty}\frac{1-\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}}$$ $$=\frac{1-0}{\sqrt{4-0}+\sqrt{4-0}}=\frac{1}{2+2}=\color{red}{\frac{1}{4}}$$
• $\dfrac{1}{2+2} = \dfrac{1}{4}$ – DeepSea Sep 28 '15 at 4:51
• After rationalizing, shouldn't it have been $4x^2-6-4x^2\color{red}{-x}$ instead of $4x^2-6-4x^2\color{red}{+x}$? – Zol Tun Kul Sep 28 '15 at 5:51
• 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})$$=4x^2-6-4x^2\color{red}{+x}$$ – Harish Chandra Rajpoot Sep 28 '15 at 6:30 • But isn't it$\sqrt{4x^2+x}$instead of$\sqrt{4x^2-x}$? – Zol Tun Kul Sep 29 '15 at 4:49 • No, because we substitute$x=-x$then$x\to +\infty$– Harish Chandra Rajpoot Sep 29 '15 at 5:11 hint: Use the identity:$\sqrt{A}-\sqrt{B} = \dfrac{A-B}{\sqrt{A}+\sqrt{B}}$, and$\sqrt{4x^2\pm6} = x\sqrt{4\pm\dfrac{6}{x^2}}$• Does the first identity have a name I can search for? – Zol Tun Kul Sep 28 '15 at 4:52 • Its a variant of the well known$A^2-B^2 = (A-B)(A+B)$. – DeepSea Sep 28 '15 at 5:09 • 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. – Zol Tun Kul Sep 28 '15 at 5:18 • You missed the$x$in front of each square root, and divide the top by$x. – DeepSea Sep 28 '15 at 5:26 \begin{align} \lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)&=\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)\cdot\frac{\sqrt{4x^2-6}+\sqrt{4x^2+x}}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\ &=\lim_{x\to-\infty}\frac{(\sqrt{4x^2-6})^2-(\sqrt{4x^2+x})^2}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\ &=\lim_{x\to-\infty}\frac{4x^2-6-(4x^2+x)}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\ &=\lim_{x\to-\infty}\frac{(-x-6)/|x|}{\sqrt{4-6/x^2}+\sqrt{4+x/x^2}}\\ &=\lim_{x\to-\infty}\frac{1+\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}\\ &=\frac{1}{\sqrt{4}+\sqrt{4}}\\ &=\frac{1}{4} \end{align} • When you divide(-x-6)/|x|$, 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. – Zol Tun Kul Sep 29 '15 at 5:06 • @ZolTunKul: In order to make manipulations inside the surds. – Ángel Mario Gallegos Sep 29 '15 at 14:16 • 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. – Zol Tun Kul Sep 30 '15 at 15:44 • @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 $$\frac{-x-6}{|x|}=\frac{-x-6}{-x}=\frac{x+6}{x}=1+\frac{6}{x}$$ – Ángel Mario Gallegos Sep 30 '15 at 15:52 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\$