<s>[INST] 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. 
 [/INST] If you know equivalents in Asymptotic analysis, that is easy:
$\;3^n+2^n\sim3^n$, $\sim5\times3^n$, hence
$$\frac{3^n+2^n}{5\times3^n+7\times2^n}\sim\frac{3^n}{5\times3^n}=\frac15.$$
</s>