# Homework Help: Limits Approaching Infinity

1. Jun 26, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
How would you take this limit? The function is:
$$lim_{x\rightarrow -∞}\frac{\sqrt{9x^{6}-x}}{x^{3}+1}$$

2. Relevant equations

3. The attempt at a solution

Uh, square root of negative infinity?

Graphing tells me that this should go to -3.

I just recalled that I can pull the 9 out to get 3root(-∞) ?

2. Jun 26, 2011

### eumyang

Nooooooo... you can't just divide by x3 when you have a square root in the numerator!

3. Jun 26, 2011

### gb7nash

You made an error with the squareroot. On the numerator, you're multiplying by 1/x^3 (which is fine). How do you proceed to move this into the squareroot?

4. Jun 26, 2011

### QuarkCharmer

We briefly touched on this idea in class Friday, and only talked about polynomials, and the proof of the way we determine horizontal asymptotes without limits.

$$lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^{6}-x}}{x^3}}{1+\frac{1}{x^3}}$$

I'm really not sure where to go from there?

5. Jun 26, 2011

### eumyang

Note that
$$x^3 = -\sqrt{x^6}$$
(since x is approaching negative infinity)

So
$$lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^{6}-x}}{x^3}}{1+\frac{1}{x^3}} = lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^{6}-x}}{-\sqrt{x^6}}}{1+\frac{1}{x^3}} = ...$$

6. Jun 26, 2011

### QuarkCharmer

I get what to do now, for the most part. Now I assume I must have some sign error. As noted above
$$x^3=\sqrt{x^6}$$
I don't understand how you got the negative there. The exponent is odd. Is it because the limit to -inf assumes x is negative?

7. Jun 26, 2011

### Char. Limit

Yes, that's right. Since x is negative, x^3 is the negative square root of x^6.

8. Jun 26, 2011

### QuarkCharmer

So, should I be dividing everything by the negative highest degree in the cases when x tends to negative infinity?

If that is the case, then the top would evaluate to -3 sure, but the denominator would be -1 after the limit, making the solution positive 3?

9. Jun 26, 2011

### Char. Limit

No, no. You're still dividing top and bottom by x^3, not -x^3. However, in the "denominator of the numerator", you're going to change x^3 to -sqrt(x6). Now, why -sqrt(x6) and not +sqrt(x6)? The reason is because x^3 is only equal to sqrt(x6) for positive x. For negative x, x^3 is equal to -sqrt(x6).

Hopefully I helped out a bit here.

10. Jun 26, 2011

### QuarkCharmer

I'm not sure I understand why only the numerators denominator is getting the negative?

11. Jun 26, 2011

### SammyS

Staff Emeritus
Because you leave the denominator's denominator as x3.

In the numerator's denominator you change x3 to $-\sqrt{x^6}\,.$

12. Jun 26, 2011

### QuarkCharmer

So essentially, the whole expression is getting it's sign changed then?

13. Jun 26, 2011

### SammyS

Staff Emeritus
Yes ... At least that's the way in appears.

Actually, the expression is less than zero (i.e., negative) for x < -1 . It's just that you need to explicitly provide a negative sign when you change x3 to $-\sqrt{x^6}$ if x < 0 .