Recent content by mileena

Thank you both. I will hopefully become a pro at this sooner than later. But it will take a lot of diligence on my part.

Thanks Enigman! I will check that site out too.

Thank you rkum99! I had forgotten about using a graphing calculator. I just bought a TI-89, so it's pretty new to me. Also, I learned there are at least two sites that will graph a function for you: This one is pretty simple to use: http://www.fooplot.com This one is a bit more...

Thanks for the tip. I am really high strung, so I get nervous very quickly. I was going to ask if the sine function has a horizontal asymptote, but then I calmed down, looked it up, and found that the sine function doesn't begin to converge on one point, as you do with an asymptote, but it goes...

Homework Statement lim x→∞ ##\frac{7x^2 + x + 11}{4 - x}##Homework Equations The Attempt at a Solution I am sorry I am posting so much. But I think I have learned two different ways to compute limits at infinity of functions: one by the math lab tutor and another by the professor, but I am...

Also, technically, Principle 1 above: "Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞." is violated here as x→∞, since the limit of the function here is ∞ but there is no horizontal asymptote. Maybe I should email my professor and ask her to clarify this. It worked...

Thank you! So I hope this is right: as x→∞, the limit is also ∞, and there is no horizontal asymptote. But as x→-∞, both the limit and horizontal asymptote is 0. For the overall function, the horizontal asymptote is y = 0. And I also will go back and check the LaTeX. I would love to...

Hi Zondrina! Yes, you interpreted the formula correctly. I didn't realize horizontal asymptotes were subject to a direction. I thought they applied to the whole function, so I saw that as x→-∞, the function approached 0. So I thought that was a horizontal asymptote (y = 0) for the function...

Homework Statement Find the limit and any horizontal asymptotes: lim 4/(e-x) [SIZE="1"]x→∞ Homework Equations Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞. Principle 2: If a limit goes to ∞ or -∞, there won't be a horizontal asymptote. The Attempt at a Solution...

Thank you. Yes, I did mess up. :approve: I know you are trying to help, but right now some things are still above my head often. As for "sgn" above, I am not even aware of that operation. Hopefully I won't have to learn it for a while; I am so overwhelmed right now!

Yes, thank you! I think that is what I did above. And I think that's where all the confusion arose in my understanding in this thread!

Ok, I think that the numerator in my problem is actually negative: √(2x2 + 1) / x But since x = -∞: √(2x2 + 1) / -|x| √(2x2 + 1) / -√X2 -√(2 + 0) -√2 So the overall answer is negative, as the denominator is positive (3 - 0 = 3).

I see that x = √x2 only if x is positive. But if x is negative, then x ≠ √x2, since the square root function returns only positive values of x. So, using the rules I have been taught, maybe the numerator is negative actually, because I am still getting a positive denominator using my...

Thank you for the encouragement! I guess I was trying to figure out why the numerator had to be divided by x2, whereas the denominator had to be divided by just x. (Normally, you would divide both the numerator and denominator by the same term, so that term equals 1. Like if you divide both...

Thank you so much vela for taking the time to do that! I looked at it closely at first to make sure I understood everything. That made things much more clear for me! I agree that √x = |x|. And that |x| = -x for negative numbers. In this case though, am I right to say that for the numerator, in...