Prove $\lim \frac{3^n + 2^n}{5\cdot3^n + 7\cdot2^n} = \frac{1}{5}$ Prove using limit definition.
$$\lim \frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} = \frac{1}{5}
$$
My try:
$$\left| {\frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} - \frac{1}{5}} \right| < \varepsilon
$$
$$ \Leftrightarrow \left| \frac{3^n + 2^n - 5\cdot 3^n - 7\cdot 2^n}{5(5\cdot3^n + 7\cdot2^n)} \right| < \varepsilon \
$$
$$ \Leftrightarrow \frac{4\cdot 3^n + 5\cdot 2^n}{5(5\cdot 3^n + 7\cdot 2^n)} < \varepsilon 
$$
I can increase numerator next and simplify like this:
$$\frac{4\cdot 3^n + 5\cdot 2^n}{5(5\cdot 3^n + 7\cdot 2^n)} < \,\frac{5\cdot 3^n + 7\cdot 2^n}{5\cdot (5\cdot 3^n + 7\cdot 2^n)} = \frac{1}{5}
$$
But in this case I'm loosing the $n$ in the dominator. 

HINT

$$
\frac{ 3^n + 2^n }{ 5 \times 3^n + 7 \times 2^n } 
= \frac{ \color{red}{1} + (2/3)^n }{ \color{red}{5} + 7 \times (2/3)^n }
$$

