Hi Steve,

I'm not sure I understand what you mean by "the numerator is dominated by the x^5 term......". The x^5 term has nothing to do with the negative sign in the denominator.

(x^10)^(1/2) is positive for all x except 0. Whereas, x^5 is positive for x > 0, and negative for x < 0. That being said, x^5 and (x^10)^(1/2) are not equal. However, since x is going to negative infinity, we can attach a negative to the latter term, and the limits will be equal.

Additionally, your notion that infinity divided by infinity is false. Infinity divided by infinity can be many different things (as you will see in some of the later lessons). Remember: Infinity is not a number, it's a concept.

Let me know if you have any further questions! Hopefully I didn't confuse you : )

Hi Evan,

Thank you for your response.

Here is the video I tried to emulate when solving the problem:

https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-continuity/modal/v/more-limits-at-infinity 

At 0:25 in the video, Sal talks about which terms will dominate. He indicates that the 9x⁷ term of the numerator will dominate, and the 3x⁷ term of the denominator will dominate. Consequently, for large values of x, i.e. x -> ∞, Sal approximates the equation as 9x⁷/3x⁷. Sal then proceeds to cancel out the x⁷ terms, leaving 9/3 or just 3 as the answer.

I attempted to use this same logic when working the following problem.

lim x -> -∞ (x⁵ + 4x²)/(x¹⁰ + 8x⁷)^.5

lim x -> -∞ (x⁵)/(x¹⁰)^.5

lim x -> -∞ x⁵/|x⁵|

lim x -> -∞ -1/1

lim x -> -∞ -1

Answer is – 1

The numerator will be negative, as it is a negative number raised to an odd power. The denominator is positive, as x is raised to an even power before the square root. Hence, to keep the denominator positive I used the absolute value operator. I then cancelled the X5 terms, leaving -1/1 or just -1.

I hope this better explains what I was attempting to do.

That being said, I believe I understand the solution that was presented.

Thank you for your help.

I found the following video that solves a similar problem in the way I approached the previously mentioned problem: https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-continuity/modal/v/limits-with-two-horizontal-asymptotes