So I have to find the horizontal asympottes of

$$ f(x)=(\sqrt{x^2+1}-|x|)(x+2) $$

Relatively straightforward. For the limit at infinity, the absolute value I turn into a positive and solve it and I get 1/2 which is correct.

But for the negative infinity portion, I get division by 0 if I simplify and "plug in" infinity.

I'm not allowed to use L'Hopital's rule on this. I rationalized it as normal but I end of getting:

$$ \lim_{x \rightarrow \infty}\frac{ x+2 }{ \sqrt{x^2+1}-x } $$

But if I got ahead and simplify that, I get get $$1/0$$ which is clearly undefined. Any ideas?

I used the assumption that I turn the absolute value negative as it goes to negative infinity. Is that wrong?


As $x\to -\infty$ we have that: $$f(x) = \left(\sqrt{x^2+1}+x\right)(x+2)= \frac{\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)(x+2)}{\sqrt{x^2+1}-x}=\frac{x+2}{\sqrt{x^2+1}-x}.$$ Thus, $$\lim_{x\to-\infty}f(x)\begin{array}[t]{l} = \lim\limits_{x\to-\infty}\dfrac{x+2}{\sqrt{x^2+1}-x}=\lim\limits_{x\to-\infty}\dfrac{x+2}{-x\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\\[3ex] =\lim\limits_{x\to-\infty}\dfrac{x}{-x\left(\sqrt{1+\frac{1}{x^2}}+1\right)}+\lim\limits_{x\to-\infty}\dfrac{2}{-x\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\\[3ex] =-\dfrac 12 +0 =-\dfrac 12. \end{array} $$

As $x\to -\infty$:

$\sqrt{x^2+1} - x=\sqrt{x^2\left(1+\frac{1}{x^2}\right)}-x =|x|\sqrt{1+\frac{1}{x^2}} - x =-x\sqrt{1+\frac{1}{x^2}}-x = -x\left(\sqrt{1+\frac{1}{x^2}}+1\right)$

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  • $\begingroup$ That is EXACTLY what I did to get 1/2 but the -1/2 part does not work when I do the same thing. $\endgroup$ – Future Math person Oct 3 '15 at 2:10
  • $\begingroup$ I ALMOST did what you did but on the denominator I factored out an x instead of a -x. Is that wrong? $\endgroup$ – Future Math person Oct 3 '15 at 2:23
  • $\begingroup$ OMG, I completely forgot about that! Thanks! $\endgroup$ – Future Math person Oct 3 '15 at 2:33
  • 1
    $\begingroup$ You are welcome! $\endgroup$ – thanasissdr Oct 3 '15 at 2:34

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