Vtyjkyuk

Given the following function:

$$f(x)= frac3x-1 sqrt3x^2-2x+1$$

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 $frac1x$.
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 $y= sqrt3$ , but according to the answers there’s another one in negative square root of three, what am I missing here?

The shortest way is with equivalents:

$3x^2-2x+1sim_pminfty3x^2$, $;3x-1sim_pminfty3x$, so
$$ frac3x-1sqrt3x^2-2x+1sim_pminftyfrac3xsqrt3x^2=sqrt3,frac x=sqrt3,operatornamesgnx. $$

When we take $;xto -infty;$, we must change sign, otherwise that thing can not go into a square root (or any even root), thus: for $;x<<0;$, we have

$$frac-frac1x-frac1xcdotfrac3x-1sqrt3x^2-2x+1=frac-3+frac1xsqrt3-frac2x+frac1x^2xrightarrow[xto-infty]frac-3sqrt3=-sqrt3$$

Remember, $;sqrtx^2=|x|;$ , and thus: if $;x<0;$ , then $;sqrtx^2=-x;$ ...

Your method works great when $x to infty$.

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

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.