What is the Limit of (2x)*[(ln(9)+1)/(ln(13x)+1)] as x approaches infinity?

[Badgerspin]

I've been having problems with evaluating this one limit. Everytime I work it out, or rework it, I keep getting infinity. It's a webwork problem for my calc 2 class, and each time I submit the answer, it tells me I'm wrong. I plugged it into wolfram as well to confirm, and it tells me infinity. So at this point, I'm at a slight loss. I did the rest of the work with no issue. Now, confusion confusion confusion.

lim(x->inf)(2x)*[(ln(9)+1)/(ln(13x)+1)]

Here's an image with the equation: http://courses.webwork.maa.org:8080/wwtmp/equations/b7/8c4810e18b1bd95df21c9f7e037d061.png

I feel like there's something I'm missing here.

Any help would be appreciated.

 

[lanedance]

is the limit as below? (click on it to see how its written in tex)
\lim_{x \to \infty} 2x \frac{ln(9) +1}{ln(13x)+1}

then
= \lim_{x \to \infty} 2(ln(9) +1) \frac{x}{ln(13x)+1}

from there you can apply L'Hopital & i agree it tends to infinity, as x increase faster than a logarithm

are you sure you started with the right expression? may be my browser but I can't view that picture

 

[mynameisfunk]

what was your answer and how did you get to it?

 

[Badgerspin]

I did similar to as the poster above did. Remove the constants and set them aside. I'm not seeing how the answer can be anything but infinity, yet the system continually tells me it's wrong.

My time expires in a few hours, and I have other work to do. I'm trying to see at this point if the professor will accept it on paper for that one problem. I can't ignore the fact that there just might be a glitch in the system.

Still, I did all of the other ones just fine and without any issues at all. This is driving me insane!

 

[Badgerspin]

is the limit as below? (click on it to see how its written in tex)
\lim_{x \to \infty} 2x \frac{ln(9) +1}{ln(13x)+1}

then
= \lim_{x \to \infty} 2(ln(9) +1) \frac{x}{ln(13x)+1}

from there you can apply L'Hopital & i agree it tends to infinity, as x increase faster than a logarithm

are you sure you started with the right expression? may be my browser but I can't view that picture

Yes, I started with the right expression. The initial expression you posted (Which by the way, I'd like to learn how to do that. It's much neater and easier to read.) looks exactly like what I was given.

Thank you for your help. I really wanted to be sure it wasn't me that was doing something wrong.

 

[lanedance]

you thinking is correct for the limit written in #2, learning to back yourself when your arguments are solid is an important part of maths, as the answer is not always known in advance

 

[Badgerspin]

Ug, please disregard. The format I was given really didn't clarify that everything attached was an exponent to (2x). A friend of mine just pointed that out to me.

I feel slightly embarrassed.

In my own defense (the images didn't work at my university, so I started with the text versions), I would like to post the verbatim version that was given to me on the site.

limx→∞(2x)ln9+1ln(13x)+1

Well, you can't win them all...

Thanks for your help. It's very much appreciated.