MHB 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

 

[Dethrone]

I don't think this requires l'hopital's rule. We have $\lim_{{n}\to{\infty}}(3/4)^{n+1}=\lim_{{n}\to{\infty}}(3/4)^n$.
Because $(3/4)^n$ is a continuous function, we can bring the limit inside to get $(3/4)^{\infty}=0$.

EDIT:

I suppose you could also do it this way:

$e^{\ln(3/4)^n}=e^{n \ln(3/4)}=e^{-n\ln(4/3)}$
Hence, $\lim_{{n}\to{\infty}}e^{-n\ln(4/3)}=e^{-\infty}=0$

 

[seal308]

I don't think this requires l'hopital's rule. We have $\lim_{{n}\to{\infty}}(3/4)^{n+1}=\lim_{{n}\to{\infty}}(3/4)^n$.
Because $(3/4)^n$ is a continuous function, we can bring the limit inside to get $(3/4)^{\infty}=0$.

I'm just learnign about sequences right now.
What does making the limit inside mean?
Do you mean something like it becomes lim(3^(n+1) / lim(4^(n+1))
and because the denominator is bigger than the numerator it goes to zero?

 

[Greg]

For any $$|x|<1,\quad\lim_{n\to\infty}x^n=0$$

 

[Dethrone]

Do you mean something like it becomes lim(3^(n+1) / lim(4^(n+1))
and because the denominator is bigger than the numerator it goes to zero?

This is essentially the intuition behind why the limit goes to zero. For example, if we take a number $c$ such that $|c|<1$, then obviously $\lim_{{n}\to{\infty}}c^{n}=0$. In this case, $c=3/4$.

There is a theorem that states for a composite function $f(g(x)$, that $\lim_{{x}\to{a}}f(g(x)=f(\lim_{{x}\to{a}}g(x))$ if $f$ is continuous at $g(a)$. In this case, we have $\lim_{{n}\to{\infty}}(3/4)^n=(3/4)^{\lim_{{n}\to{\infty}}n}=(3/4)^{\infty}=0$.

 

[seal308]

This is essentially the intuition behind why the limit goes to zero. For example, if we take a number $c$ such that $|c|<1$, then obviously $c^{\infty}=0$. In this case, $c=3/4$.

There is a theorem that states for a composite function $f(g(x)$, that $\lim_{{x}\to{a}}f(g(x)=f(\lim_{{x}\to{a}}g(x))$ if $f$ is continuous at $g(a)$. In this case, we have $\lim_{{n}\to{\infty}}(3/4)^n=(3/4)^{\lim_{{n}\to{\infty}}n}=(3/4)^{\infty}=0$.

Wow thanks, that makes so much sense.
Ty everyone.

 

[Nono713]

$c^{\infty}$ is not really defined notation. It's much clearer to write that:
$$\lim_{n \to \infty} c^n = 0$$
Which can be proven for $\lvert c \rvert < 1$ through e.g. sequences and stuff.

 

[Dethrone]

$c^{\infty}$ is not really defined notation. It's much clearer to write that:
$$\lim_{n \to \infty} c^n = 0$$
Which can be proven for $\lvert c \rvert < 1$ through e.g. sequences and stuff.

I thought that would be an issue, but thought the OP would understand it better if written in a more explicit fashion. Anyway, I have edited it to reflect this.

 

[Prove It]

The OP needs to realize that when multiplying an amount by a small fraction (i.e. with a size less than 1) you get a SMALLER number as a result. So continually multiplying by a small fraction will continually result in a smaller amount.