Limit as x approaches negative infinity (sqrt(9x^6 - x))/(x^3+1) = ? The answer is -3 but I don't understand

Can't the answer be positive three?

So the way I did it, since you get an indeterminate form of +∞/-∞ when you solve directly is that I divided the numerator and denominator by the highest exponent power of x (which is x³). So I got:

√(9x^6 - x) ∙ (1/x³)
--------------------------
(x³ + 1) ∙ (1/x³)

But since the numerator is under a square root sign, to simplify, you need to change the x³ into a square root, too.

You might think you can just substitute x³ = √(x^6), but for negative numbers of x (which is what you will have, since you are taking the limit when x --> -∞) x³ will be a negative number.

In other words:

x³ = -√(x^6) ≠ √(x^6)

So you get:

√(9x^6 - x) ∙ (-1/√(x^6))
--------------------------
(x³ + 1) ∙ (1/x³)

This simplifies to:

-√(9 - (1/x^5))
--------------------
1 + 1/x³

Since 1/x^5 and 1/x³ both go to zero as x goes to negative infinity, you get:

-√(9 - 0)
------------
1 + 0

= -3

I hope that helps!