Is it possible to prove that the limit, as n goes to infinity, of the sequence $a_n = 3^n/7^n$ is $0$? Intuitively this seems to be true, but is there a way to actually prove it instead of just saying "the denominator gets bigger faster"? I'm not sure that the sequence can be modeled by a function, but even in doing that and using l'Hopital's rule I didn't get 0.  
 A: This is a pretty standard and basic result related to sequences.

Theorem: Let $x\in\mathbb {R} $ then the sequence $\{a_n\} $ defined by $a_n=x^n$
  
  
*
  
*diverges to $\infty$ if $x>1$.
  
*converges to $1$ if $x=1$.
  
*converges to $0$ if $|x|<1$.
  
*oscillates finitely with values $1,-1$ if $x=-1$.
  
*oscillates infinitely if $x<-1$.
  

Your case is about $x=3/7$ and then from 3rd bullet point above the desired limit is $0$.
The proof of the theorem above is not that difficult and is a very instructive one. I give below a simple proof for the first case which also implies the third case (which applies here).
If $x>1$ then we can write $$x^n=(1+(x-1))^n>1+n(x-1)$$ via binomial theorem. Now the RHS of the inequality tends to $\infty $ and therefore so does the LHS.
Now we deal with the third case. If $x=0$ the result is obvious. So let $0<|x|<1$ so that $y=1/|x|>1$ and therefore from last paragraph $y^n\to\infty $. And then $|x^n|=1/y^n\to 0$. Since $$-|x^n|\leq x^n\leq |x^n|$$ it follows by squeeze theorem that $x^n\to 0$.
A: The argument relies on the completeness property of $\mathbb{R}$, or at least the Archimedean property. A proof in $\mathbb{R}$ using the standard completeness axiom:
Let $a_n=\left(\frac37\right)^n$. This sequence is bounded below by $0$, so it has a greatest lower bound $L$. It's a decreasing sequence, so $L$ is its limit.  Now, the sequence $b_n=\frac37a_n$ has limit $\frac37L$. But at the same time, $b_n=a_{n+1}$, so $L=\lim_n a_n=\lim_n b_n=\frac37L$ and $L=0$. Done.
Proving that the limit is zero if it exists? That's just limit arithmetic. Proving that the limit exists, on the other hand, requires more.
A: To prove that $\lim_{n\to\infty}\frac{3^n}{7^n}=0$, you have to show that there exists $M \in \mathbb{N}$ such that
$$n  \geq M \implies  \big(\frac{3}{7} \big)^n=\big| \,\frac{3^n}{7^n} - \color{red}{0} \, \big| < \epsilon$$
Since $\ln(x)$ is strictly increasing, we can apply it on both sides of the inequality above to get
$$n\cdot\ln(\frac{3}{7})=\ln\big(\frac{3}{7} \big)^n < \ln(\epsilon)$$
Because $\ln(\frac{3}{7})<0$, you get $$n > \frac{\epsilon}{\ln(3/7)}$$
Therefore, setting $M=\bigg\lceil\frac{\Large{\epsilon}}{\ln(3/7)}\bigg\rceil$ will work where $\lceil \cdot \rceil$ denotes the ceiling function, i.e. the first integer that is bigger than or equal to what is inside the brackets.
A: (1). The Archimedean property: If $r\in  \Bbb R^+$ there exists $m\in \Bbb N$ with $m>r.$
(2). Corollary: If $y,w \in \Bbb R^+$ there exists $m\in \Bbb N$ with $ym>w.$ Proof: Let $r=w/y$ and apply (1).
(3). If $1<1+y\in \Bbb R$ and $0<w\in \Bbb R^+$ there exists $m\in \Bbb N$ with $(1+y)^m>w.$
Proof: For $m\in \Bbb N$ we have $(1+y)^m\ge 1+ym,$  by induction on $m,$ or by the Binomial Theorem (which itself is proved by induction on $m$). Now by (2), take $m\in \Bbb N$ with $ym>w.$ Then $(1+y)^m\ge 1+ym>ym>w.$
(4). If $0<x<1$ let $x=1/(1+y).$ Now, given $\epsilon >0,$ let $w=1/\epsilon .$ Then $y,w\in \Bbb R^+$ so by (3),  take $m\in \Bbb N$ with $(1+y)^m>w>0.$ So $$x^m=1/(1+y)^m <1/w=\epsilon.$$ 
Now for any $n\in \Bbb N,$ if  $n\ge m$  then $n-m\ge 0$ and $0<x<1$ so $0<x^{n-m}\le 1.$ Therefore   $$m\le n\in \Bbb N\implies 0<x^n=x^{n-m}x^m\le x^m<\epsilon.$$
Remark: The Archimedean property is a direct consequence of the definition of the structure $\Bbb R.$
