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

 

[Ishuda]

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

First we know that the (3/4)ⁿ⁺¹ is positive for every finite n so
\(\displaystyle 0\, \le\, (\frac{3}{4})^{n+1}\)
Now suppose there were some positive number a greater than zero such that
\(\displaystyle \underset{n\, \to\, \infty}{lim}\, (\frac{3}{4})^{n+1}\, = a\)
What does that mean? Now note that
\(\displaystyle (\frac{3}{4})^{n+2}\, =\, \frac{3}{4}\, (\frac{3}{4})^{n+1}\, \lt\, (\frac{3}{4})^{n+1}\)

 

[Steven G]

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

If you keep multiplying a positive number less than 1 forever it will keep getting smaller. That is EXACTLY what Ishuda was pointing out.

If you want to prove this more rigorously (good for you!) then let y = (3/4)^(n+1).
So ln(y) =(n+1)ln(3/4).

Then lim as n goes to infinity of ln(y) = lim as n goes to infinity of (n+1)ln(3/4)=negative infinity (note than ln(3/4)<0). Now if ln(y) goes to -inf as x goes to inf, then y goes to 0 as x goes to inf.

 

[seal308]

Yes I get it now.
The denominator moves faster than the numerator so it gets closer and closer to 0.

 

[Steven G]

Yes I get it now.
The denominator moves faster than the numerator so it gets closer and closer to 0.

But that is not always true. Consider n^2/(n+1)^2 as n goes to infinity. The denominator moves faster than the numerator, yet the limit is 1, NOT 0.

Did you understand my proof above?. Understand that inf/inf can end up being any real number including =/- inf.