Not able to find the horizontal asymptote

Question

find horizontal asymptote

$y=(x^2+x)^{1/2}-(x^2-1)^{1/2}$

I have tried multiplying both numerator and denominator with $(x^2+x)^{1/2}+(x^2-1)^{1/2}$ but don't know how to proceed

Answer 1

$$\eqalign{y &=(x^2+x)^{1/2}-(x^2-1)^{1/2}\cr &=\frac{((x^2+x)^{1/2}-(x^2-1)^{1/2})((x^2+x)^{1/2}+(x^2-1)^{1/2})}{(x^2+x)^{1/2}+(x^2-1)^{1/2}}\cr &=\frac{x+1}{(x^2+x)^{1/2}+(x^2-1)^{1/2}}\cr &=\frac{1+\frac1x}{(1+\frac1x)^{1/2}+(1-\frac1{x^2})^{1/2}}\cr &\to\frac12\quad\hbox{as}\ x\to\infty\ .\cr}$$

Answer 2

You got pretty close

$$y = \sqrt{x^2 + x} - \sqrt{x^2-1}$$ $$y= \frac{x+1}{\sqrt{x^2+x}+\sqrt{x^2-1}}$$

Now we have to look at the limit as $x$ approaches $\pm \infty$

$$\lim_{x \to +\infty}\frac{x+1}{\sqrt{x^2+x}+\sqrt{x^2-1}} \approx \frac{x}{\sqrt {x^2} + \sqrt{x^2}}\approx \frac{x}{2|x|} = \frac{1}{2}$$ $$\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2+x}+\sqrt{x^2-1}} \approx \frac{x}{\sqrt {x^2} + \sqrt{x^2}}\approx \frac{x}{2|x|} = -\frac{1}{2}$$