How to find the other asymptote of $y=\sqrt{x^2+x}$

Question

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?

Answer 1

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.

Answer 2

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?