Finding the Constant k for a Solvable Limit Problem

[stangman]

Im stuck on one problem in particular that I cannot figure out how to start but once I get past the first step I can figure it out. The problem is...
lim as x-->infinity (3^kx + 6) / (3^2x + 4) and you must find a value of constant k such that the limit exists. I was thinking that k might be 2 so that the entire terms could cancel out.

 

[radou]

Try dividing the fraction with 3^(kx).

 

[stangman]

now when you divide 3^2x by 3^kx, that brings you to 3^2x-kx correct? Now do you factor out x to make it 3^x(2-k) in which case the entire equation would become 1 / 3^x(2-k) after substituting infinity into the other terms but I don't understand what to do with this last term.

 

[radou]

Okay, start with $$lim_{x \rightarrow \infty} \frac{3^{kx}+6}{3^{2x}+4}=lim_{x \rightarrow \infty} (\frac{3^{kx}}{3^{2x}+4}+\frac{6}{3^{2x}+4})=lim_{x \rightarrow \infty}\frac{3^{kx}}{3^{2x}+4}+lim_{x \rightarrow \infty} \frac{6}{3^{2x}+4}=lim_{x \rightarrow \infty}\frac{3^{kx}}{3^{2x}+4}$$. Now divide with 3^(kx) and factor out x, as mentioned above.

 

[stangman]

after following that I came out to have 1 / 3^x(2-k) ... is that what you mean by factoring out x? However I still don't understand what the value of k would be if i still have x in the equation. Maybe I am just not seeing something..

 

[radou]

after following that I came out to have 1 / 3^x(2-k) ... is that what you mean by factoring out x? However I still don't understand what the value of k would be if i still have x in the equation. Maybe I am just not seeing something..

You now have $$lim_{x \rightarrow \infty} \frac{1}{3^{x(2-k)}+\frac{4}{3^{kx}}}$$. All you have to do now is to see if there will be a limit for different values of k. The interesting cases are k = 2, k > 2 and k < 2.

 

[dextercioby]

I advise you to use the "\" before "lim" in order to look better.

$$\lim_{x\rightarrow \infty}$$

C?

Daniel.

 

[radou]

I advise you to use the "\" before "lim" in order to look better.

$$\lim_{x\rightarrow \infty}$$

Thanks... I just don't understand how to get the $$x \rightarrow \infty$$ below the limit. The source code is the same, I don't get it. Anyway, off topic. :)

 

[stangman]

so now if k=2 the limit is 1, if k > 2 then the limit is infinity, if k < 2 then the limit is 0, did I solve that right or what am I missing now?

 

[stangman]

does it mean that the limit exists at k < or = to 2?

 

[stangman]

Yeah I caught my mistake for k < 2 right after I posted it but how is the limit not 1 for k = 2, when 2 is put into 3^x(2-k) that would make it 3^0 ultimately which equals 1, and 4/3^kx becomes 0

 

[radou]

does it mean that the limit exists at k < or = to 2?

Yes, for k < 2 it exists, and it equals zero. Further on, it exists for k = 2, and it equals 1. Sorry, I messed something up. :)

 

[stangman]

Alright I finally understand it now, thank you very much for the help

 

[stangman]

ok one more question, if the problem was the exact same but this time approach negative infinity, I did out the problem and would the limit now exist at k > or = to 2?

 

[radou]

If $$x \rightarrow -\infty$$, for k < 0, the limit is infinity, for k = 0 it equals 1/4, and for k > 0 it equals zero.

 

[stangman]

may I just ask why you use 0 instead of 2?

 

[radou]

may I just ask why you use 0 instead of 2?

Because for k = 0 'something changed'.

 

[stangman]

you still use the same method of simplifying the equation as we did for the first question right?

 

[radou]

Yes. In our case, it was all about bringing the function (whose limit is of interest to us) into a suitable form to apply the simple limit $$\lim_{x\rightarrow\infty}\frac{1}{x}=0$$.

 

[stangman]

true true, however in the first one we were able to cancel out the other side of the formula of 6/3^2x + 4 because of that simple limit. However now you cannot because when negative infinity is put into its spot it becomes undef

 

[radou]

Right, I forgot about the right side, whose limit is 3/2 for $$x\rightarrow -\infty$$. So, $$\lim_{x\rightarrow -\infty}\frac{3^{kx}+6}{3^{2x}+4} = \left\{\begin{array}{ccc}\infty, k<0 \\ \frac{7}{4}, k = 0\\\frac{3}{2}, k>0 \end{array}\right$$.

 

[stangman]

alright then that finally all makes sense, I just ask how or why you chose zero for k to test, is there a particular reason, I know you said something changed but I just don't understand that part.

 

[radou]

true true, however in the first one we were able to cancel out the other side of the formula of 6/3^2x + 4 because of that simple limit. However now you cannot because when negative infinity is put into its spot it becomes undef

Because k was a factor in the exponent, and k = 0 turned the term to 1. The point is: try to find values which simplify the expression.

 

[Jacobpm64]

stangman, lol.. just curious, but I was looking through my book, and I saw this exact problem... I'm using a book called Calculus:Single Variable (Hughes-Hallett), lol just wondering if you're using the same book, and if so.. what school are you going to?

 

[stangman]

yea I'm using the same book as you, I am in uconn, u?

 

[Jacobpm64]

northwestern state university .. in louisiana.. I thought maybe we'd be in the same class or something and not even know.. would be hilarious.. but.. guess not.. ha.. take it easy then :P

 

[stangman]

lol yea thought so to, I also some other of your threads and they were problems that I had done to so I was gettin suspicious myself

 

[Jacobpm64]

i sent you a private message.. check it.

 

[istevenson]

the correct answer should be this:
0 < k < 2 or k = 2 or k =0.