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

• seal308
In summary, the limit of (3/4)^(n+1) as n approaches infinity is 0. This can be seen by writing the expression as e^(n+1)ln(3/4) and noting that the ln is negative, causing the whole expression to approach zero as n goes to infinity. L'Hopital's rule is not necessary.
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

seal308 said:
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.

mfb said:
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?

seal308 said:
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

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.

Last edited:

## 1. What is the limit of (3/4)^(n+1) as n approaches infinity?

The limit of (3/4)^(n+1) as n approaches infinity is 0. This means that as n gets larger and larger, the value of (3/4)^(n+1) gets closer and closer to 0.

## 2. How do you calculate the limit of (3/4)^(n+1) as n approaches infinity?

To calculate the limit of (3/4)^(n+1) as n approaches infinity, you can use the formula lim(x->inf) f(x) = lim(x->inf) f(x)/g(x) if the limit exists and g(x) is nonzero at x=a. In this case, we can simplify the expression to (3/4)^n, and since 3/4 is less than 1, the limit is 0.

## 3. Is the limit of (3/4)^(n+1) as n approaches infinity equal to 0?

Yes, the limit of (3/4)^(n+1) as n approaches infinity is equal to 0. This means that as n gets larger and larger, the value of (3/4)^(n+1) gets closer and closer to 0.

## 4. How does the value of n affect the limit of (3/4)^(n+1) as n approaches infinity?

The value of n does not affect the limit of (3/4)^(n+1) as n approaches infinity. The limit will always be 0, regardless of the value of n.

## 5. What does it mean when the limit of (3/4)^(n+1) as n approaches infinity is 0?

When the limit of (3/4)^(n+1) as n approaches infinity is 0, it means that as n gets larger and larger, the value of (3/4)^(n+1) will get closer and closer to 0. This can also be interpreted as the function approaching 0 as n approaches infinity.

• Calculus and Beyond Homework Help
Replies
17
Views
1K
• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
20
Views
2K
• Calculus and Beyond Homework Help
Replies
24
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
565
• Calculus and Beyond Homework Help
Replies
17
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
945
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
948