limit to infinity with variable as exponent

[alex_dc]

So this is the only question in the chapter that I haven't been able to figure out. Honestly, I'm not even sure where to start. I know the answer is 27, but that doesn't help me solve the problem.

Problem:

((3^(x+1)) - (2^(x+4)) / ((3^(x-2)) + (2^(x-1)) + 6)

as x --> infinity

The only place I can think to start is to apply the identities (a^(x+y)) = (a^x) * (a^y), and (a^(x-y)) = (a^x) / (a^y), but (after simplifying and playing around with that a bit) I honestly have no idea where to go from there.

Any help is appreciated! Thanks!

 

[Deleted member 4993]

So this is the only question in the chapter that I haven't been able to figure out. Honestly, I'm not even sure where to start. I know the answer is 27, but that doesn't help me solve the problem.

Problem:

((3^(x+1)) - (2^(x+4)) / ((3^(x-2)) + (2^(x-1)) + 6)

as x --> infinity

The only place I can think to start is to apply the identities (a^(x+y)) = (a^x) * (a^y), and (a^(x-y)) = (a^x) / (a^y), but (after simplifying and playing around with that a bit) I honestly have no idea where to go from there.

Any help is appreciated! Thanks!

\(\displaystyle \lim_{x\to\infty}\left[\frac{3^{x+1}-2^{x+4}}{3^{x-2}+2^{x+1}+6}\right]\)

dividing the numerator and denominator by 3[sup:1fufpftn]x[/sup:1fufpftn], we get,

\(\displaystyle \lim_{x\to\infty}\left[\frac{3^{1}-2^{4} (\frac{2}{3})^x}{3^{-2}+2^{1}(\frac{2}{3})^x+\frac{6}{3^x}}\right]\)

Now continue......

 

[alex_dc]

Oh, duh. I totally overcomplicated a simple problem. So we have:

(2/3)^x = (2^x)/(3^x)

divide top and bottom by 2^x and we have:

1/((3/2)^x)

which as x --> infinity, 1/((3/2)^x) --> 0

So we have, as x --> infinity

3 --> 3

2^4((2/3)^x) --> 0

3^-2 --> 1/9

2((2/3)^x) --> 0

6/(3^x) --> 0

Which means:

3 - (2^4)((2/3)^x) --> 3

(3^-2) + (2)((2/3)^x) + (6)(3^x) --> 1/9

so:

(3 - (2^4)((2/3)^x)) / ((3^-2) + (2)((2/3)^x) + (6)(3^x)) --> 3/(1/9)

3/(1/9) = (3*9)/1 = 27

Thank you.