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

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

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mfb
Mentor
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.

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
Mentor
Am I overthinking this.
Yes.
seal308 said:
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.

OK I'll use that logic then. thx

Mark44
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.

• seal308
I see thanks again.

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.

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