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

  • Thread starter seal308
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  • #1
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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
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
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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.
 
  • #5
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OK I'll use that logic then. thx
 
  • #6
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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.
 
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  • #7
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I see thanks again.
 
  • #8
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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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