Find the limit of an exponential sequence I can't find the limit of the sequence
$b_n= \frac{2^n+3^n}{3\cdot2^{n+1}+2\cdot3^{n-1}}$
I managed to go up until:
$\frac{1+(3/2)^n}{6+2/3\cdot(3/2)^n}$
but I am still stuck since I  get infinity\infinity
Is it possible to find the limit for that function or maybe there is no limit?
Thanks in advance
 A: I suppose that the limit is considered for $n \rightarrow \infty$. Your idea is not bad but I would suggest going this way 
$$ b_n = \frac{2^n+3^n}{3\cdot2^{n+1}+2\cdot3^{n-1} }= \frac{3^n((2/3)^n+1)}{3^{n-1}(3\cdot4\cdot(2/3)^{n-1}+2)} = 3 \frac{(2/3)^n+1}{3\cdot4\cdot(2/3)^{n-1}+2} \ .$$ 
Considering the expression on the right for $n \rightarrow \infty$ should lead you quickly to the result as you can go with the limit up to the numerator and down to the denominator (the final hint maybe to think about how the powers of a positive constant smaller then $1$ works). 
A: $$b_n-\frac32=-\frac{8\cdot 2^n}{3\cdot2^{n+1}+2\cdot3^{n-1} }=-\frac{8}{6+\left(\dfrac32\right)^{n-1} }\to0.$$
A: I would split the fraction into the sum of two with the common denominator.  The first term I would find an equivalent fraction by dividing numerator and denominator by $$2^{n-1}$$.  Looking at the limit now of that first term it will be 0 since $$(3/2)^{n-1}$$ approaches infinity.  Using the same approach, the second term can be rewritten by dividing the numerator and denominator by $$3^{n-1}$$ but now this term's limit approaches 3/2 since $$(2/3)^{n-1}$$ approaches 0.  
So the result is 3/2.  Notice that there are many ways to answer this.
A: $$\frac{2^n+3^n}{3\cdot2^{n+1}+2\cdot3^{n-1}}=\frac{3\left(\left(\dfrac23\right)^n+1\right)}{3\cdot4\left(\dfrac23\right)^{n-1} +2}\to\frac{0+3}{0+2}\to\frac 32$$ 
