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.  

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 $x^n\to 0$.
