How do you take the limit of a function approaching negative infinity?

[QuarkCharmer]

Homework Statement

How would you take this limit? The function is:
$$lim_{x\rightarrow -∞}\frac{\sqrt{9x^{6}-x}}{x^{3}+1}$$

Homework Equations

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(-∞) ?

 

[eumyang]

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

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

 

[gb7nash]

Homework Statement

How would you take this limit? The function is:
$$lim_{x\rightarrow -∞}\frac{\sqrt{9x^{6}-x}}{x^{3}+1}$$

Homework Equations

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(-∞) ?

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?

 

[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?

 

[eumyang]

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?

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

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

 

[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?

 

[Char. Limit]

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?

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

 

[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?

 

[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.

 

[QuarkCharmer]

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

 

[SammyS]

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

Because you leave the denominator's denominator as x³.

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

 

[QuarkCharmer]

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

 

[SammyS]

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

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 .