I'm trying to calculate the slant asymptotes of the function $\sqrt{x^2+2x+2}$. I've found out that the gradients of the asymptotes are 1 for $x\rightarrow+\infty$ and -1 for $x\rightarrow-\infty$. I've also found out that the constant of the positive asymptote is 1. Intuitively, I know the constant of the negative asymptote is -1, but I'm struggling to show it through calculation. I need to evaluate this to find it:

$$\lim_{x\rightarrow-\infty} \sqrt{x^2+2x+2} + x $$

without using l'Hôpital's rule (for the purposes of the assignment I'm not supposed to know how to use it.) I have tried rationalizing the numerator using the conjugate but I just end up with an undefined value.


Hint: $x^2+2x+2=(x+1)^2+1,$ and $\sqrt{(\text{stuff})^2}=|\text{stuff}|,$ so for $x\ne-1,$ we have $$\sqrt{x^2+2x+2}=\sqrt{(x+1)^2+1}=|x+1|\cdot\sqrt{1+\frac1{(x+1)^2}}.$$

If you want to evaluate that limit directly, it sounds like you got off to the right start by noting that $\sqrt{x^2+2x+2}$ is never equal to $x$ for real $x,$ so we can certainly say that $$\begin{align}\sqrt{x^2+2x+2}+x &= \left(\sqrt{x^2+2x+2}+x\right)\cdot\frac{\sqrt{x^2+2x+2}-x}{\sqrt{x^2+2x+2}-x}\\ &= \frac{\left(\sqrt{x^2+2x+2}+x\right)\left(\sqrt{x^2+2x+2}-x\right)}{\sqrt{x^2+2x+2}-x}\\ &= \frac{\left(\sqrt{x^2+2x+2}\right)^2-x^2}{\sqrt{x^2+2x+2}-x}\\ &= \frac{2x+2}{\sqrt{x^2+2x+2}-x}\\ &= \frac{2x}{\sqrt{x^2+2x+2}-x}+\frac{2}{\sqrt{x^2+2x+2}-x}.\end{align}$$ Now, noting that $$\lim_{x\to\infty}\sqrt{x^2+2x+2}-x=+\infty,$$ it follows that $$\lim_{x\to-\infty}\sqrt{x^2+2x+2}+x=\lim_{x\to-\infty}\frac{2x}{\sqrt{x^2+2x+2}-x}=2\lim_{x\to-\infty}\cfrac1{\frac1x\sqrt{x^2+2x+2}-1}.$$ But for $x<0,$ we have $$\begin{align}\frac1x\sqrt{x^2+2x+2} &= -\frac1{|x|}\sqrt{x^2+2x+2}\\ &= -\frac1{\sqrt{x^2}}\sqrt{x^2+2x+2}\\ &= -\sqrt{1+\frac2x+\frac2{x^2}}.\end{align}$$ Can you take it from there?

  • $\begingroup$ Sorry, I made a typo! It should be +2. $\endgroup$ – Christiaan Swanepoel Aug 21 '13 at 5:32
  • $\begingroup$ Edited. Can you take it from there? $\endgroup$ – Cameron Buie Aug 21 '13 at 5:37
  • $\begingroup$ So as $x\rightarrow\infty$, $f(x)\rightarrow |x+1|$? $\endgroup$ – Christiaan Swanepoel Aug 21 '13 at 5:49
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    $\begingroup$ Oh, for Pete's sake! Never mind, I've got an idea. I'll add to my answer shortly. $\endgroup$ – Cameron Buie Aug 21 '13 at 6:12
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    $\begingroup$ Answer expanded. You should be able to evaluate the limit now. $\endgroup$ – Cameron Buie Aug 21 '13 at 6:27

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