Homework Help: 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

courtrigrad

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

6. Oct 17, 2006

HallsofIvy

How, exactly, are you defining ax for a< 0?

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