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I'm having troubles with finding oblique asymptotes of a function $$ x-3\sqrt{4x^2+5} $$ This is what I did : $$ x-3\sqrt{x^2(4+\frac{5}{x^2})} $$ So I thought I could assume that $$ f(x) \approx x-3|x| $$ So I proceeded to finding asymptotes for $$ -2x / 4x $$ The results are $$ \frac {-2x} {x} = -2 $$ $$ \frac {4x} {x} = 4 $$ $$ -2x+2x = 0 $$ $$ 4x -4x = 0 $$ $$ y= -2x / y=4x $$ However Wolfram Alpha says the asymptote is $$ y=-5x $$ https://www.wolframalpha.com/input/?i=asymptote+x-3sqrt(4x%5E2%2B5)

What I possibly don't understand about finding oblique asymptotes for functions involving absolute values?

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    In the very beginning, you should factor out $(2x)^2$ from the radical. – 2017-02-18
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    Shouldn't it be $f(x)=x-6|x|$? (I know nothing about asymptotes, but it seems $\sqrt{4x^2+5} \approx 2x$). – 2017-02-18
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    How can I be sure what number should be factored out? $$ x-3\sqrt{x^2(4+\frac{5}{x^2})} | x-3\sqrt{2x^2(2+\frac{5}{2x^2})} | x-3\sqrt{4x^2(1+\frac{5}{4x^2})} $$ I think all of those representations are right, but they give different asymptotes? – 2017-02-18
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    Ok, it seems that square root out of 4 determined what should be factored out. – 2017-02-18

3 Answers 3

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We can write $$y=x-3|x|\sqrt{4+\frac{5}{x^2}}$$

So for large positive $x$, we have $$y\simeq x-3(x)(2)\implies y\simeq-5x$$

And for large negative $x$, we have $$y\simeq x-3(-x)(2)\implies y\simeq 7x$$

Note that there must be two asymptotes since the curve is a hyperbola.

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    It should be y=7x for me, too. I think Wolfram Alpha fails to calculate asymptote from the "other" side. – 2017-02-18
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    Yes, so it seems. There may be some way of obtaining the other asymptote from WA but I don't know what that is. – 2017-02-18
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    This is not a mathematical proof. How can you be sure that $y+5x$ and $y-7x$ tend to $0$? – 2017-02-18
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    @DavidQuinn try replacing x with -x and looking at that asymptote – 2017-02-18
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You fail to incorporate the radical part. Notice that

$$\sqrt{4+\frac5{x^2}}\to\sqrt4=2$$

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Starting form.
$\quad y=x-3\sqrt{4x^2+5}$
Isolate radical term.
$\quad y-x=3\sqrt{4x^2+5}$
Square both sides.
$\quad (y-x)^2=9(4x^2+5)$
$\quad y^2-2xy+x^2=36x^2+45$
Implicit form.
$\quad 35x^2+2xy-y^2+45=0$
Substitute $y=mx+b.$
$\quad 35x^2+2x(mx+b)-(mx+b)^2+45=0$
$\quad 35x^2+2mx^2+2bx-m^2x^2-2mbx-b^2+45=0$
Extract coëfficients of $x^2,\quad x;$
equate them to $0;$
and solve for $m,\quad b.$
$\quad \left\{\begin{array}{l} 35+2m-m^2=0\\ -2mb=0 \end{array}\right.$

$\begin{array}{c|ccccc} \text{case}&\{m\mid b\}&\to&mx+b=y&\to&\text{asymptote}\\ \hline 1&\{7\mid 0\}&\to&7x+0=y&\to&7x=y\\ 2&\{-5\mid 0\}&\to&-5x+0=y&\to&-5x=y \end{array}$