Need help with solving/simplifying a difficult infinite sum/limit

Question

I am trying to figure out how to solve for this infinite sum: $$\lim _{x\to\infty} \frac {3^x 2^0+3^{x-1} 2^1+3^{x-2} 2^2+\cdots+3^0 2^x}{2 (3^x+3^{x-1}+3^{x-2}+\cdots+3^0)}.$$ The "..."s mean to continue the infinite sum until you reach the value on the right. I believe there would be a way to rewrite this sum purely algebraically, but I need help figuring out how to do so. By examining the first few instances of this sum (by ignoring the limit and setting x to a finite number), I believe the sum either converges to 1 or something larger, but I'm not sure of it's specific value.

Answer 1

$2(3^0+3^1+3^2+...+3^{x-2}+3^{x-1}+3^x)$

$=3^0+3^0+3^1+3^1+3^2+3^2+...+3^{x-1}+3^{x-1}+3^x+3^x$

$=3^1-3^0+3^2-3^1+3^3-3^2+3^4-3^3+...+3^x-3^{x-1}+3^{x+1}-3^x$

$=-1-3^{x+1}$

You can prove that $3^n+3^n=3^{n+1}-3^n$ for all natural numbers $n$.

For the numerator, I prove the equality in this question, so I will apply $a^n + a^{n-1}b+a^{n-2}b^2+\ldots+ab^{n-1}+b^n=\dfrac{a^{n+1}-b^{n+1}}{a-b}$ here:

${3^x 2^0+3^{x-1} 2^1+3^{x-2} 2^2+\cdots+3^0 2^x}$

$={3^{x+1}-2^{x+1}}$ because $3-2=1$.

Then $\lim _{x\to\infty} \frac {3^x 2^0+3^{x-1} 2^1+3^{x-2} 2^2+\cdots+3^0 2^x}{2 (3^x+3^{x-1}+3^{x-2}+\cdots+3^0)}=\lim _{x\to\infty} -\frac{3^{x+1}+1}{3^{x+1}-2^{x+1}}$, which you can continue from there, and as you say the fact that the expression converges to $1$, which is true.