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

[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

 

[mfb]

lim = ( ln(3) * 1 * 3^(n+1) ) / ( ln(4) * 1 * 4^(n+1) )

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.

 

[seal308]

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.

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?

 

[Mark44]

Am I overthinking this.

Yes.

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

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

 

[seal308]

OK I'll use that logic then. thx

 

[Mark44]

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.

 

[seal308]

I see thanks again.

 

[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.