How to find the limit of this series I was trying to figure out the limit of the function below
$a_n = \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}$
The answer to the question is $\frac{1}{8}$ but when I divide through by $3^{n+3}$ (because it has the highest power) I cant figure out how they got that answer, Im really stuck has anyone go any hints 
 A: $$a_n = \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}=$$
$$\frac{(1/2)4^n+(27)3^n}{(9)3^n+(4)4^n}=$$
$$\frac {(1/2)+(27)(3/4)^n}{(9)(3/4)^n+(4)}\to 1/8$$
A: For such ratios, one should guess first what grows faster. Normally, $a^{k.n\pm l}$, for $a>1$, $k$ and $l$ integers, behaves as $(a^k)^n$. 
So with similar fractions, it is often useful to factor both the numerator and the denominator with the biggest $(a_i^{k_i})^{n_i}$ of each. 
Finally, you end up with something like 
$$\frac{\nu^{n}}{\delta^{n}}\frac{N+o(1)}{D+o(1)}$$
which give you a global behavior. In your case, you can get $\nu=\delta=4$, and $N/D=1/8$.
A: Look for hidden powers:
$a_n 
= \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}
= \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+2^{2n+2}}
$.
The $3^n$ terms are negligible
for large $n$,
so
$a_n 
\approx \frac{2^{2n-1}}{2^{2n+2}}
=2^{(2n-1)-(2n+2)}
=2^{-3}
=\frac18
$.
To check,
$\begin{array}\\
a_n-\dfrac18
&= \dfrac{2^{2n-1}+3^{n+3}}{3^{n+2}+2^{2n+2}}-\dfrac18\\
&= \dfrac{8(2^{2n-1}+3^{n+3})-(3^{n+2}+2^{2n+2})}{8(3^{n+2}+2^{2n+2})}\\
&= \dfrac{2^{2n+2}+8\cdot 3^{n+3})-3^{n+2}-2^{2n+2}}{8(3^{n+2}+2^{2n+2})}\\
&= \dfrac{8\cdot 3^{n+3}-3^{n+2}}{8(3^{n+2}+2^{2n+2})}
\qquad\text{The } 2^{2n+2} \text{ is cancelled out}\\
\text{so}\\
|a_n-\dfrac18|
&= \dfrac{|8\cdot 3^{n+3}-3^{n+2}|}{|8(3^{n+2}+2^{2n+2})|}\\
&\lt \dfrac{|8\cdot 3^{n+3}|}{2^{2n}|8(3^3(3/4)^{n}+2^{2})|}\\
&= \left(\dfrac{3}{4}\right)^n\dfrac{|8\cdot 3^{3}|}{|8(3^3+4)|}\\
&\lt \left(\dfrac{3}{4}\right)^n\dfrac{216}{248}\\
&\to 0\\
\end{array}
$
