simple convergence test $\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}$

i am pulling my hair out in solving this problem. i know, it is a stupid question but i am not that good at maths, and many thanks for any help

$\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}$

my steps are: $\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n} = \frac{2^{n}2+3^{n}3}{2^n+3^n} = help = ? $ i know, it diverges as the denominator grows faster than the nominator, but the calculation is killing me now

The main idea here and in similar problems is to factor out the terms not with the biggest exponents but with the biggest bases. That way you will create terms of the form $(a/b)^n$ where $a<b$ that converge to $0$. $$\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}=\lim_{n \to \infty} \frac{3^{n+1}((\frac23)^{n+1}+1)}{3^n((\frac23)^n+1)}=\lim_{n \to \infty} 3\frac{(\frac23)^{n+1}+1}{(\frac23)^n+1}=3 $$ because $(\frac{2}{3})^n\to 0$

Your first step is good. Now divide top and bottom by $3^n$.