# Finding $\lim \frac{3^n}{4^n}$

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

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.

• A standard $\epsilon$-$N$ argument should suffice. – JMoravitz Feb 20 '20 at 21:14
• Hint: Write the expression as $\left( \frac 34\right)^n$. Note: you should make it clear that you mean $\lim_{n\to \infty}$. – lulu Feb 20 '20 at 21:15
• @JMoravitz What's that? – Segmentation fault Feb 20 '20 at 21:15
• Well, $\frac{3^n}{4^n} = (\frac{3}{4})^n$ and $\frac{3}{4} < 1$... – PrudiiArca Feb 20 '20 at 21:15
• @InterstellarProbe Why though? – Segmentation fault Feb 20 '20 at 21:16

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.

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 $$s^n \to 0\quad\text{as } n \to \infty, \text{when } 0 Hint: use $$t=1/s$$.

• +1 too same reason – Satyendra Feb 20 '20 at 22:01

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.

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

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

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 $$0$$.

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