Flipping signs for limits at negative infinity

[farleyknight]

Homework Statement

$\lim_{x \to -\infty} x + \sqrt{x^2 + 6x}$

Homework Equations

The Attempt at a Solution

Previous attempt was guessing it was $\infty$, but I see now my flaw and the actual answer is -3. Somewhere else on the web, might have been this forum, it was said that one could flip the sign and get

$\lim_{x \to -\infty} x + \sqrt{x^2 + 6x} = \lim_{x \to \infty} -x + \sqrt{x^2 + 6x}$

Which I can see intuitively, since the what is under the radical would be positive either way, which implies that the sign only need be flipped for x. However, is there a generalized proof that includes any number of polynomials and roots, for this fact?

Thanks,
- Farley

 

[Dick]

What you are doing is substituting -u for x. Then as x->-infinity, u->+infinity. But x+sqrt(x^2+6x) turns into -u+sqrt(u^2-6u), doesn't it?

 

[farleyknight]

Yeah, that's probably a mistake.. But that does explain it. Thanks.