Why is the limit $\lim_{x\to-\infty} \frac{2x}{\sqrt{1+x^2}}-\arctan x=\pi/2-2$ and not $\pi/2 + 2$?

Question

Why is the limit $$\lim_{x\to-\infty} \frac{2x}{\sqrt{1+x^2}}-\arctan x=\pi/2-2$$ and not $\pi/2 + 2$?

Evaluating I calculate the "2" to be positive but wolfram alpha says its negative. I've made the mistake before but can't figure out what the reason is.

Answer 1

$$\lim_{x\to-\infty} \frac{2x}{\sqrt{1+x^2}}=\lim_{t\to\infty} \frac{-2t}{\sqrt{1+t^2}}=\lim_{t\to\infty} \frac{-2}{\sqrt{1+\frac 1{t^2}}}=-2$$

Quite possibly, the error might have occurred when you would have directly taken x inside the square root.

Answer 2

With asymptotic equivalents, it's still more obvious: we have $\:1+x^2\sim_{\pm\infty}x^2$, so $\:\sqrt{1+x^2}\sim_{\pm\infty}|x|$ and therefore $$\frac{2x}{\sqrt{1+x^2}}\sim_{-\infty}\frac{2x}{|x|}=2\,\operatorname{sgn} x=-2.$$

Answer 3

$\lim _{x\to - \infty} \arctan(x) =-\frac{\pi}{2} $ and

$ \lim _{x\to - \infty} \frac{2x}{\sqrt{1+x^2}} = \lim _{x\to - \infty} - \frac{2}{\sqrt{1+\frac{1}{x^2} }} = - 2$

Finally :

$\lim _{x\to - \infty} \frac{2x}{\sqrt{x^2 +1}}-\arctan(x) =\frac{\pi}{2} - 2$

Do not forget that: $\lim _{x\to - \infty} \sqrt{x^2} = \lim _{x\to - \infty} - x$