Strategy for the Limit: $\lim_{n\to\infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n} $ I do not understand how to properly solve this limit: 
$$
\lim_{n\to\infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}
$$
I thought of breaking it up: 
$$
\lim_{n\to\infty} \frac{2^{n+1}}{2^n+3^n} +\lim_{n\to\infty} \frac{3^{n+1}}{2^n+3^n}
$$
But I do not see how this will allow me to use any of the limit rules to reduce. I know that the series converges though. 
Thanks. 
 A: The rule of the dominant term : always divide by the most dominant term on top and bottom, and see where things go. (The dominant term is the algebraic expression growing the fastest, usually detected by observation).
For example, here the most dominant term is $3^{n+1}$. So divide top and bottom by $3^{n+1}$ :
$$
\frac{2^{n+1} + 3^{n+1}}{2^n + 3^n} = \frac{\left(\frac{2}{3}\right)^{n+1} + 1}{\frac 13\left(\frac 23\right)^n + \frac 13}
$$
The limit of the numerator is $1$ and the denominator is $\frac 13$ as $n \to \infty$, and thus the desired limit is their quotient i.e. $3$. The limits of top and bottom are easy to calculate since we have the power of $\frac 23 < 1$ in both expressions, which goes to $0$ as $n \to \infty$. 
Dividing  by the most dominant term allows you to create a numerator and denominator whose limits are easy to calculate.
A: Factor out $\frac{2^n}{2^n}$ from your first term and factor $\frac{3^n}{3^n}$ from your second term.  If you do this correctly you should find that the answer is $3$.
A: Since the 3s dominate,
I would expect the limit
to be 3.
Check:
$\begin{array}\\
\dfrac{2^{n+1}+3^{n+1}}{2^n+3^n}-3
&=\dfrac{2^{n+1}+3^{n+1}-3(2^n+3^n)}{2^n+3^n}\\
&=\dfrac{2^{n+1}+3^{n+1}-3\cdot 2^n-3^{n+1}}{2^n+3^n}\\
&=\dfrac{2^{n+1}-3\cdot 2^n}{2^n+3^n}\\
&=\dfrac{-2^n}{2^n+3^n}\\
&\to 0
\qquad\text{since } 2^n/3^n \to 0\\
\end{array}
$
