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$$\lim_{n \to \infty} \frac{3^n}{4^n}$$

I know the limit is zero because the denominator grows faster than the numerator in this case... although I still get infinity over infinity.

How do I "show" that the limit is zero? L'Hopital's rule is redundant in this case and doing $\lim e^{n\ln(3/4)}$ doesn't help.

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  • $\begingroup$ A standard $\epsilon$-$N$ argument should suffice. $\endgroup$ – JMoravitz Feb 20 '20 at 21:14
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    $\begingroup$ Hint: Write the expression as $\left( \frac 34\right)^n$. Note: you should make it clear that you mean $\lim_{n\to \infty}$. $\endgroup$ – lulu Feb 20 '20 at 21:15
  • $\begingroup$ @JMoravitz What's that? $\endgroup$ – Segmentation fault Feb 20 '20 at 21:15
  • $\begingroup$ Well, $\frac{3^n}{4^n} = (\frac{3}{4})^n$ and $\frac{3}{4} < 1$... $\endgroup$ – PrudiiArca Feb 20 '20 at 21:15
  • $\begingroup$ @InterstellarProbe Why though? $\endgroup$ – Segmentation fault Feb 20 '20 at 21:16
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I don't think Lhopital's rule is redundant.

Call the limit $L$, which exists because the sequence is bounded and decreasing.

The derivative of $b^x$ is $\ln bb^x$.

We get $L=\ln3/\ln4L\implies L=0$.


Also, $e^{n\ln(3/4)}\to0$, since $\ln(3/4)$ is negative.

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One textbook proceeds like this.

(1) Bernoulli's inequality: $$ (1+x)^n \ge 1+nx\qquad\text{when } x > 0, n \in \mathbb N $$ Hint: induction.

(2) Use (1) to show $$ t^n \to \infty\quad\text{as } n \to \infty, \text{when } t > 1 $$ Hint: use $1+x=t$.

(3) Use (2) to show $$ s^n \to 0\quad\text{as } n \to \infty, \text{when } 0<s<1 $$ Hint: use $t=1/s$.

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  • $\begingroup$ +1 too same reason $\endgroup$ – Satyendra Feb 20 '20 at 22:01
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The sequence $x_{n}=\left(\frac{3}{4}\right)^n$ has recursive definition $x_0=1, x_{n+1}=\frac{3}{4}x_n.$

$\{x_n\}$ is decreasing and bounded below by zero. So it has a limit, $L.$

But $$L=\lim_{n\to\infty} x_{n+1} = \frac{3}{4}\lim_{n\to\infty} x_{n}=\frac{3}{4}L.$$

So $L=0.$

So you really only need that a decreasing sequence bound below has a limit, and simple properties of limits.

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Perhaps like this:

$\dfrac{1}{(4/3)^n}=\dfrac{1}{(1+1/3)^n}<$

$\dfrac{1}{1+n/3}<\dfrac{1}{n/3} =3/n.$

Used: $(1+x)^n \gt 1+xn$, for $x>0$, $n$ positive integer

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  • $\begingroup$ +1 Nice answer with Bernouilli's inequality. $\endgroup$ – Satyendra Feb 20 '20 at 21:59
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    $\begingroup$ LostinSpace.Thanks. $\endgroup$ – Peter Szilas Feb 20 '20 at 22:04
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HINT

Note $3^n/4^n = (3/4)^n$ so let $\epsilon > 0$. Can you find $N$ such that $(3/4)^n < \epsilon$ for all $n > N$?

This would show that the positive sequence $(3/4)^n$ gets arbitrarily small, which is the definition of convergence to $0$.

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  • $\begingroup$ No, how do I find it? $\endgroup$ – Segmentation fault Feb 20 '20 at 21:16
  • $\begingroup$ @SilenceOnTheWire $(3/4)^n < \epsilon \iff n > \frac{\ln \epsilon}{\ln (3/4)}$... $\endgroup$ – gt6989b Feb 20 '20 at 21:17
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    $\begingroup$ @gt6989b $\ln(3/4)<0$, so it flips the direction of the inequality: $$\left(\dfrac{3}{4}\right)^n < \epsilon \Longrightarrow n > \dfrac{\ln \epsilon}{\ln 3 - \ln 4}$$ $\endgroup$ – InterstellarProbe Feb 20 '20 at 21:20
  • $\begingroup$ @InterstellarProbe +1, thanks, fixed mine, but yours is also formatted better, don't delete it... $\endgroup$ – gt6989b Feb 20 '20 at 21:22

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