Flipping signs for limits at negative infinity

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SUMMARY

The limit of the expression \(\lim_{x \to -\infty} x + \sqrt{x^2 + 6x}\) evaluates to -3. The discussion highlights a method of flipping the sign by substituting \(-u\) for \(x\), transforming the limit into \(\lim_{x \to \infty} -x + \sqrt{x^2 + 6x}\). This approach is intuitive as the expression under the square root remains positive. A generalized proof for flipping signs in limits involving polynomials and roots is sought.

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

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