# Constant k and limits

1. Sep 20, 2006

### 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.

2. Sep 20, 2006

Try dividing the fraction with 3^(kx).

3. Sep 20, 2006

### 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 dont understand what to do with this last term.

4. Sep 20, 2006

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.

Last edited: Sep 20, 2006
5. Sep 20, 2006

### 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 dont understand what the value of k would be if i still have x in the equation. Maybe im just not seeing something..

6. Sep 20, 2006

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.

7. Sep 20, 2006

### dextercioby

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

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

C?

Daniel.

8. Sep 20, 2006

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. :)

Last edited: Sep 20, 2006
9. Sep 20, 2006

### 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?

Last edited: Sep 20, 2006
10. Sep 20, 2006

### stangman

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

11. Sep 20, 2006

### 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

12. Sep 20, 2006

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. :)

13. Sep 20, 2006

### stangman

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

14. Sep 20, 2006

### 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?

15. Sep 20, 2006

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.

16. Sep 20, 2006

### stangman

17. Sep 20, 2006

Because for k = 0 'something changed'.

18. Sep 20, 2006

### stangman

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

19. Sep 20, 2006

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 $$\lim_{x\rightarrow\infty}\frac{1}{x}=0$$.