# Limit of a^x

1. Oct 16, 2006

### Siegfried

Hi,

I was doing some exercises on limits when I stumbled upon the following questions:
(limits are always x->infinity)

lim (2^x+1)/(3^x+1)

2nd one:
lim (-9/8)^x

The first one should be zero (3^x expands faster than 2^x), and the 2nd one doesn't seem to exist (according to maple and the solutions , I suppose it has something to do with the graph being all chopped up and not knowing if x->infinity is even or odd?).
Is there a way to find these answers trough calculation instead of uncertain logic :tongue: ?
Another exercise similar to the first one:
lim (8/9)^x

2. Oct 16, 2006

### quasar987

$$a^x = \exp(\ln a^x) = \exp(x \ln a) = (e^x)^{\ln a}$$

you go from there.

3. Oct 16, 2006

### Siegfried

So lim(8/9)^x = lim e^(x*ln(8/9))
ln(8/9) (or ln (8 ) -ln(9)) is smaller than 0, resulting in a negative value for x*(ln(8/9))
So the result would be something of this form:
lim e^(-x*...)
or
lim 1/e^(x*...)
going towards 0
Am I correct?

Now, the limit of (-9/8)^x can't be calculated this way unless we use irrational numbers ( ln(a) = ln(-a)+Pi*I for a<0 ) , right? But is this the proof that the limit doesn't exist? or is there another reason?

4. Oct 16, 2006

### quasar987

My hint was aimed at solving the first limit.

For the second one, you could show that the sequence {(-9/8)^2n} goes to +infinity and the sequence {(-9/8)^(2n+1)} goes to -infinity.

5. Oct 16, 2006

for the second one you could also show that it is a geometric sequence with $$|r| > 1$$.