Does $\lim_{x \to - \infty} \left(\frac{\pi}{2} + \arctan{x} \right) \cdot x = - \infty$?

Question

Does $$\lim_{x \to - \infty} \left(\frac{\pi}{2} + \arctan{x} \right) \cdot x = - \infty$$?

My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?

Answer 1

The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:

1. It is sloppy. Mathematics is rigorous for a reason. Mathematical statements must be well defined, and have a single meaning. In your case, what is "something"? Is an elephant times negative infinity equal to negative infinity?
2. Even if you formalize the sentence, to something like "For all $\alpha$, the limit $\lim_{x\to-\infty}\alpha\cdot x=-\infty$, the statement remains false, since it is not true for any $\alpha\leq 0$.
3. The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$\lim_{x\to-\infty} \frac{1}{x}\cdot x=-\infty,$$ since we have "something" times "negative infinity". Clearly, this is nonsence - the limit is obviously equal to $1$

---

The limit you must calculate is a limit of a product of two numbers, one tends to $0$, the other to $-\infty$. Such a limit is often best approached using L'Hospital's rules.

Answer 2

No. By L'Hopital's Rule $\lim \frac {\frac {\pi} 2+\arctan\, x} {1/x}=\lim \frac {1/(1+x^{2})} {-1/x^{2}}=-1$.

Answer 3

Let $-1/x=h$

$$\lim_{h\to0^+}\dfrac{\dfrac\pi2-\arctan\dfrac1h}{-h}=-\lim_{...}\dfrac{\arctan h}h=-1$$

See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?