<s>[INST] Limit of sequence-term not always positive Let the sequence $a_n=\frac{3^n+2(-4)^n}{(-4)^n-2^n}$.
We have $$\frac{3^n+2(-4)^n}{(-4)^n-2^n}=\frac{3^n+2(-1)^n 4^n}{(-1)^n 4^n-2^n}=\frac{(-1)^n 4^n \left( 2+ \frac{3^n}{(-1)^n 4^n}\right)}{(-1)^n 4^n \left( {1-\frac{2^n}{(-1)^n 4^n}}\right)}$$
In this case the ineqaulity $-1 \leq (-1)^n \leq 1 \Rightarrow -4^n \leq (-1)^n 4^n \leq 4^n$ does not help, since we cannot have an inequality with these terms as denominators, because we don't know if $(-1)^n 4^n$ is positive or negative. Right?
How else can we we compute the limit of the sequence?
 [/INST] You have 
\begin{align}
\frac{3^n+2(-4)^n}{(-4)^n-2^n}
&=\frac{3^n}{(-4)^n-2^n}+\frac{2(-4)^n}{(-4)^n-2^n}\\ \ \\
&=\frac{3^n}{4^n((-1)^n-(1/2))^n}+\frac{2}{1-2^n/(-4^n)}\\ \ \\
&\to 0+2=2
\end{align}
</s>