$$\lim_{n \to \infty} \frac{3^n}{4^n}$$
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.
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 $t=1/s$.
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
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.
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 $0$.