Asymptotes of a function with a square root

Question

I'd like to ask a question about what can I possibly do wrong with determining asymptotes of the function

$$x \mapsto x-2\sqrt{x^2+1} $$

OK, so when it comes to vertical asymptotes function, we can't have any because domain of the function is the set of all real numbers.

Now, I'm trying to determine horizontal asymptote.

$$ lim_{x\to \pm \infty} \frac{(x-2\sqrt{x^2+1})(x-2\sqrt{x^2+1})} {(x-2\sqrt{x^2+1})} = \frac {-3x^2-4}{3x+2} $$

It's an indeterminate form, so I'm using l'Hospital rule. That makes it $$ \frac{-6x}{3} = \infty $$

So, horizontal asymptote doesn't exist either.

Now, I'm determining oblique asymptotes

$$ lim_{x\to \pm \infty} \frac {\frac{-3x^2-4}{3x+2}}{x} = \frac {-3x^2-4}{3x^2+2x} = -1 $$ $$ a=-1 $$ $$ lim_{x\to \pm \infty} \frac{-3x^2-4}{3x+2} + x = \frac{-3x^2-4+x(3x+2)}{3x+2}=\frac{-3x^2-4+3x^2+2x}{3x+2}=\frac{-4+2x}{3x+2}=\frac{2}{3} $$ $$ b= \frac{2}{3} $$

so, in my opinion oblique asymptote should be: $$ y=-x+\frac{2}{3} $$

But Wolfram Alpha gives different answer : enter link description here

What mistake am I doing?

Answer 1

What is wrong is that the computation is that, contrary to what is claimed, $$\frac{\left(x-2\sqrt{x^2+1}\right)\left(x-2\sqrt{x^2+1}\right)} {\left(x-2\sqrt{x^2+1}\right)} \neq \frac {-3x^2-4}{3x+2}.$$ For one, when manipulating this way, one wants to multiply the numerator and denominator by the conjugate of the original radical expression, namely $$\frac{x\color{red}{+}2\sqrt{x^2+1}} {x\color{red}{+}2\sqrt{x^2+1}}$$ (note the change in sign.). In any case, this is not the only problem, as even when the correct sign is used, the expression for the denominator is incorrect.

In any case, there is an easier way to proceed:

Hint We can rewrite the function as $$x \mapsto x - 2 |x| \sqrt{1 + \frac{1}{x^2}}.$$

Additional hint For $x > 0$, this is$$x \mapsto x - 2 x \sqrt{1 + \frac{1}{x^2}} = x \left(1 - 2 \sqrt{1 + \frac{1}{x^2}}\right).$$ As $x \to +\infty$, the radical quantity approaches $1$, so this curve has the line $y = -x$ as an (oblique) asymptote. How does this computation chance when looking for an asymptote line as $x \to -\infty$? (Note that the linked WolframAlpha answer does not give the second asymptote that arises this way!)

Answer 2

The simplest here is to use Taylor's expansion, after the substitution $t=\dfrac1x$ (so $t\to 0^+$ or $0^-$). It will give you in one shot the asymptote, if it exists, and the position of the curve w.r.t. its asymptote.

So we can rewrite \begin{align}x-2\sqrt{x^2+1}&=\frac1t-2\sqrt{\frac1{t^2}+1}=\frac1t-\frac1{\lvert t\rvert}\sqrt{1+t^2}=\frac1t-\frac1{\lvert t\rvert}\Bigl(1+\dfrac{t^2}{2}+o(t^2)\Bigr)\\ &=\frac1t-\frac1{\lvert t\rvert}+\frac12\lvert t\rvert+o(t)=x -\lvert x\rvert+\frac1{2\,\lvert x\rvert}+o\Bigl(\frac1x\Bigr)\\ &=\begin{cases}\dfrac1{2 x}+o\Bigl(\dfrac1x\Bigr)&\text{if }\;x\to+\infty,\\2x+\dfrac1{2\,\lvert x\rvert}+o\Bigl(\dfrac1x\Bigr)&\text{if }\;x\to-\infty.\end{cases} \end{align} Thus, when $x\to+\infty$, we have the horizontal asymptote $\;y=0$, and the curve is above it if $x$ is large enough.

When when $x\to-\infty$, we have the oblique asymptote $\;y=2x$, and the curve is above it if $\vert x\rvert$ is large enough.

Answer 3

You can reason on this problem by means of the asymptotic behavior of the function (see https://en.wikipedia.org/wiki/Asymptotic_analysis).

For large $x$, we have

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

so that for positive values the function tends to $-x$, and for negative ones, to $3x$.

Then to get the intercept,

$$\lim_{x\to\infty }f(x)-(-x)=\lim_{x\to\infty }2(x-\sqrt{x^2+1})=\lim_{x\to\infty }2\frac{x^2-(\sqrt{x^2+1})^2}{x+\sqrt{x^2+1}}=0$$

and

$$\lim_{x\to-\infty }f(x)-3x=-\lim_{x\to-\infty }2(x+\sqrt{x^2+1})=-\lim_{x\to-\infty }2\frac{x^2-(\sqrt{x^2+1})^2}{x-\sqrt{x^2+1}}=0.$$

Hence the asymptotes $y=-x$ and $y=3x$.