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

-
Hint: divide each term by $3^{n+1}$. –  David Mitra Dec 15 '12 at 16:36

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 $(\frac{2}{3})^n\to 0$

-
Great, thanks a lot. i was just thinking in some totally wrong way.. –  doniyor Dec 15 '12 at 16:41
@doniyor: $(\frac{2}{3})^n\to 0$ because $|2/3|<1$. –  B. S. Dec 15 '12 at 16:42
@BabakSorouh yes, exactly. thanks –  doniyor Dec 15 '12 at 16:42
Your first step is good. Now divide top and bottom by $3^n$.
@doniyor: Or factor $3^{n+1}$ in deno. and factor $3^n$ in nomi. –  B. S. Dec 15 '12 at 16:40
The same strategy is useful for $\frac{2n^5+4n^3-1000n}{3n^5-99n^4-n^2+17}$, and many others, such as replacing top and bottom by their square roots. –  André Nicolas Dec 15 '12 at 16:51