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The task is to find the asymptotes of $y=\sqrt{x^2+x}$.

I first calculated the limits to infinity and found that $\lim_{x \to \pm}y= \infty$. Next, to find $m_{1,2}$: $$m_1=\lim_{x \to +\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_1$$

and $$m_2=\lim_{x \to -\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_2$$

Now to find $c_{1,2}$: $$c_1=\lim_{x \to +\infty}y-mx={\sqrt{x^2+x} - x}=\frac{(\sqrt{x^2+x} - x)({\sqrt{x^2+x} +x})}{{\sqrt{x^2+x} +x}}=\frac{x}{{\sqrt{x^2+x} +x}}= \frac{x}{x+{|x|\sqrt{1+\frac{1}{x}}}} \to \frac{x}{x+{|x|}} \approx \frac{1}{1+\frac{|x|}{x}}=\frac{1}{2} $$ $$c_2=\frac{1}{0}?$$

From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=\frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $\frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-\frac{1}{2}(2x+1)$.

How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?

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    $\begingroup$ $\sqrt{x^2}=-x$ if $x<0$. $\endgroup$ – egreg Dec 19 '18 at 16:54
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You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $x\to \pm\infty$ if $f(x)=mx+q+o(1)$. Now, as $|x|\to +\infty$, $$\sqrt{x^2+x}=|x|\sqrt{1+1/x}=|x|\left(1+\frac{1}{2x}+o(1/x)\right)\\ =|x|+\frac{|x|}{2x}+o(1).$$ Hence if $x>0$ then $$\sqrt{x^2+x}=x+\frac{1}{2}+o(1)$$ and for $x<0$ $$\sqrt{x^2+x}=-x-\frac{1}{2}+o(1).$$ What may we conclude?

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  • $\begingroup$ Well, since you have used asymptotic expansions, I think this means that if when $x\to \pm\infty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach. $\endgroup$ – E.Nole Dec 19 '18 at 16:49
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    $\begingroup$ Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes $\endgroup$ – Robert Z Dec 19 '18 at 18:31
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You have the wrong value for $m_2.$

If $x \leq -1,$ then $\frac{\sqrt{x^2+x}}{x} \neq \sqrt{1+ \frac{1}{x}}.$ Do you see why?

Once you have fixed this error, the rest of your calculations should work OK.

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