Limits of exponential functions

[bodensee9]

Hi,

can I do the following?

If I am asked to find as lim x -> inf of (3^n/2^n), can I do:

3^n = e^(n*ln3)
2^n = e^(n*ln2)

assume C = lim (3^n/2^n).

C = lim e^(n*ln3)/e^(n*ln2)
ln C = lim ln [(exp(n*ln3)/(exp(n*ln2)]
In C = lim ln(exp(n*ln3)) - ln(exp(n*ln2))
ln C = lim n*ln3 - n*ln2
ln C = lim n*(ln3 - ln2)

The right hand side is clearly inf, and hence ln C = inf, which would mean that C = inf?
Though wouldn't C = 0 also give you infinity? So how would I know which one it is?

Thanks.

 

[Dick]

ln(0) isn't infinity. It's undefined. n*(ln3-ln2) is clearly unbounded. So C=exp(n*(ln3-ln2)) is unbounded.

 

[bodensee9]

Thanks!

 

[HallsofIvy]

Hi,

can I do the following?

If I am asked to find as lim x -> inf of (3^n/2^n), can I do:

3^n = e^(n*ln3)
2^n = e^(n*ln2)

assume C = lim (3^n/2^n).

C = lim e^(n*ln3)/e^(n*ln2)
ln C = lim ln [(exp(n*ln3)/(exp(n*ln2)]
In C = lim ln(exp(n*ln3)) - ln(exp(n*ln2))
ln C = lim n*ln3 - n*ln2
ln C = lim n*(ln3 - ln2)

The right hand side is clearly inf, and hence ln C = inf, which would mean that C = inf?
Though wouldn't C = 0 also give you infinity? So how would I know which one it is?

Thanks.

Why do it that way?
$$\frac{3^n}{2^n}= \left(\frac{3}{2}\right)^n$$
and that, since 3/2> 1, clearly increases without bound as x increases.

On the other hand,
$$\frac{2^n}{3^n}= \left(\frac{2}{3}\right)^n$$
since 2/3< 1, clearly goes to 0.

Doing it the "hard way", if y= 2ⁿ/3ⁿ, then ln(y)= ln(2ⁿ)- ln(3ⁿ)= n(ln(2)- ln(3)). ln(2)- ln(3) is negative so, as n goes to infinity, n(ln(2)- ln(3)) goes to negative infinity and, as Dick said, ln(0)= negative infinity.