Homework Help: Limit of (3/4)^(n+1) as n approaches infinity

1. May 11, 2015

seal308

Hello,

I was just wondering how to solve this limit: Limit of (3/4)^(n+1) as n approaches infinity

My attempt:
(3/4)^(n+1) = (3^ (n+1) ) / (4 ^ (n+1))
Top goes to infinity and bottom goes to infinity to use l'hopital rule.
lim = ( ln(3) * 1 * 3^(n+1) ) / ( ln(4) * 1 * 4^(n+1) )

But the correct answer is apparently 0.
Not sure if i'm totally wrong or if I'm just halfway there.

Thanks

2. May 11, 2015

Staff: Mentor

You are not done here. Both numerator and denominator still go to infinity, and you didn't gain anything.

There are more direct approaches that work.

3. May 11, 2015

seal308

Am I overthinking this.
Is it as simple as the denominator is larger and gets bigger faster and numerator making the whole thing = 0 for that reason?

4. May 11, 2015

Staff: Mentor

Yes.
Something like that, but the "whole thing" never equals zero. It gets arbitrarily close to zero, though.

5. May 11, 2015

seal308

OK I'll use that logic then. thx

6. May 11, 2015

Staff: Mentor

If a is any number in the interval (-1, 1), then $\lim_{n \to \infty}a^n = 0$.

BTW, please don't delete the three parts of the homework template. They are there for a reason; namely, to help you organize your problem and the work you've done.

7. May 11, 2015

seal308

I see thanks again.

8. May 11, 2015

fourier jr

if you write (3/4)^(n+1) as $e^{(n+1)\ln{(\frac{3}{4})}}$ maybe it will be easier to see why it goes to zero since the ln is negative & you don't need l'hopital's rule either.

Last edited: May 11, 2015