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Given the following function:

$$f(x)= \frac{3x-1}{ \sqrt{3x^2-2x+1}}$$

I want to find the horizontal asymptotes of it, I let the function’s values go up to infinity and down go negative infinity in order to examine its behavior for growth and diminishing. For calculating it neatly I’ve multiplied both the numerator and the denominator by $\frac{1}{x}$. Then squared it and took a root of it at the denominator in order to get it under the original square root. After doing that and examine what would happen I found one horizontal asymptote at $y= \sqrt{3}$ , but according to the answers there’s another one in negative square root of three, what am I missing here?

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  • $\begingroup$ It's great that you showed your work, but would you mind showing the exact steps? $\endgroup$ – Toby Mak Sep 2 '18 at 11:38
  • $\begingroup$ @TobyMak it’ll be a pain in the arse writing it all through phone Latex :( the thing is that the numerator strived to 3 and the denominator to square root of three in both attempts to strive $\endgroup$ – Ozk Sep 2 '18 at 11:40
  • $\begingroup$ You’re asking others to spend their time helping you. It’s only fair that you spend more of your own time, too. $\endgroup$ – amd Sep 2 '18 at 20:45
  • $\begingroup$ @amd well that’s really a nonsensical thing to say.. It’s not that I didn’t try to convey my way of thinking at all. Not only that I did try, people understood my attempts completely by what I gave so that they granted me assistance based on it. Not only that this comment is nonsensical it’s rather futile because it isn’t helpful at all $\endgroup$ – Ozk Sep 3 '18 at 4:12
  • $\begingroup$ Not at all. Just responding to your whining about having to do a bit of extra typing to help out someone who’s willing to help you. $\endgroup$ – amd Sep 3 '18 at 4:34
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The shortest way is with equivalents:

$3x^2-2x+1\sim_{\pm\infty}3x^2$, $\;3x-1\sim_{\pm\infty}3x$, so $$ \frac{3x-1}{\sqrt{3x^2-2x+1}}\sim_{\pm\infty}\frac{3x}{\sqrt{3x^2}}=\sqrt3\,\frac x{|x|}=\sqrt3\,\operatorname{sgn}x. $$

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  • $\begingroup$ What’s that super super complicated but efficient method? $\endgroup$ – Ozk Sep 2 '18 at 11:46
  • $\begingroup$ It's not complicated. It uses the basics of asymptotic calculus (big O, little o, equivalents functions, &c.) $\endgroup$ – Bernard Sep 2 '18 at 11:48
  • $\begingroup$ But it doesn’t really make it any simpler because I have to evaluate the lim of x over absolute value of x in both cases $\endgroup$ – Ozk Sep 2 '18 at 17:51
  • $\begingroup$ @Ozk: This quotient is the sign function, as I wrote it. It is constant, equal to $1$ on $\mathbf R_+$, to $-1$ on $\mathbf R_-$. I don't think this limit is so hard… $\endgroup$ – Bernard Sep 2 '18 at 18:32
  • $\begingroup$ Looked up for its graph, that’s very simple indeed. $\endgroup$ – Ozk Sep 2 '18 at 19:40
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When we take $\;x\to -\infty\;$, we must change sign, otherwise that thing can not go into a square root (or any even root), thus: for $\;x<<0\;$, we have

$$\frac{-\frac1x}{-\frac1x}\cdot\frac{3x-1}{\sqrt{3x^2-2x+1}}=\frac{-3+\frac1x}{\sqrt{3-\frac2x+\frac1{x^2}}}\xrightarrow[x\to-\infty]{}\frac{-3}{\sqrt3}=-\sqrt3$$

Remember, $\;\sqrt{x^2}=|x|\;$ , and thus: if $\;x<0\;$ , then $\;\sqrt{x^2}=-x\;$ ...

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Your method works great when $x \to \infty$.

When $x\to -\infty$, you have to be a bit more careful. For $x\to -\infty$, we have eventually $x < 0$, so that $x = -|x| = -\sqrt{x^2}$, giving rise to the other horizontal asymptote.

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  • $\begingroup$ By that you mean that when x is approaching negative infinity then 1 over x approaching zero from the negative side? By the way didn’t really understand that transformation with the absolute value $\endgroup$ – Ozk Sep 2 '18 at 11:43
  • $\begingroup$ @Ozk I just wanted to point out that, for negative $x$, you don't have $x = \sqrt{x^2}$, but rather $x = -\sqrt{x^2}$. Take for example $x = -1$. Then $\sqrt{x^2} = \sqrt{(-1)^2} = \sqrt{1} = 1$, so we need the minus sign in front. $\endgroup$ – MisterRiemann Sep 2 '18 at 11:44
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Note that the bottom is always positive but the top changes sign as your $x$ goes over $x=1/3$

Thus on the positive side you get a posives limit and on the negative side you get a negative limit.

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  • $\begingroup$ You’re right, but I don’t fully understand it, X approaches positive and negative infinity, why does its behavior around one third matters? $\endgroup$ – Ozk Sep 2 '18 at 19:44
  • $\begingroup$ Other than that, the sign kinda cuts off when I divide by x $\endgroup$ – Ozk Sep 2 '18 at 19:46

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