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I'm dealing with this problem where $f(x) = (\sqrt{x^2+1} - |x|)(x+2)$

I don't think it has a vertical asymptote as no part of the equation can ever be undefined and the square root of $x^2+1$ is always positive.

To find horizontal asymptotes of rational functions I would use limits, but although this function converges at $\frac{1}{2}$ at infinity and $\frac{-1}{2}$ at negative infinity (or so it appears to me), I'm not sure if that would be regarded as an asymptote since it's not a rational function or a function that has asymptotes.

Also, if to calculate the horizontal asymptotes I need to use limits at infinities, could I get some help as to how to do that for this function without the use of L'Hopital's rule?

Thanks!

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  • $\begingroup$ It the graph indeed converges to 1/2 for x going to infinity, then you found a horizontal asymptote right there. Proving is of course a different story , but you could start by using a conjugate $\endgroup$ – imranfat Oct 9 '15 at 2:34
  • $\begingroup$ I tried using a conjugate, but it leads me nowhere. $\endgroup$ – Daniel Waleniak Oct 9 '15 at 2:50
  • $\begingroup$ See answer below, WW1 used a conjugate $\endgroup$ – imranfat Oct 9 '15 at 11:58
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$$\lim_{x \to \infty}f(x) =\lim _{x \to 0^+} \left(\sqrt{x^{-2}+1} - \left| \frac 1x \right |\right) \left(\frac 1x+2\right)$$

$$=lim _{x \to 0^+} \frac{(\sqrt{x^{2}+1} - 1 )(2x+1)}{|x|x } $$ Use the fact that $ (\sqrt{x^{2}+1} -1 )(\sqrt{x^{2}+1} +1 )=x^2 $ $$\lim_{x \to \infty}f(x)=lim _{x \to 0^+} \frac{ (2x+1)}{(\sqrt{x^{2}+1} +1 )} =\frac 12 $$

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  • $\begingroup$ Why did you let the limit as x approaches infinity equal the limit as x approaches 0 from the right? $\endgroup$ – Daniel Waleniak Oct 9 '15 at 4:03
  • $\begingroup$ let $u=\frac1x$ --- then $\lim_{x \to \infty}f(x)$ =$\lim_{u \to 0^+}F(\frac 1u)$ --- and $\lim_{x \to -\infty}f(x)$ =$\lim_{u \to 0^-}F(\frac 1u)$ $\endgroup$ – WW1 Oct 9 '15 at 5:45
  • $\begingroup$ I decided to do it similarly but still at infinity and divided by $|x|$ which gave me the same answer. Thank you. $\endgroup$ – Daniel Waleniak Oct 9 '15 at 18:36

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