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?
Best Answer
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 $$x \mapsto x - 2 |x| \sqrt{1 + \frac{1}{x^2}}.$$