How to find the limit of this series

Question

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

Answer 1

$$a_n = \frac{2^{2n-1}+3^{n+3}}{3^{n+2}+4^{n+1}}=$$

$$\frac{(1/2)4^n+(27)3^n}{(9)3^n+(4)4^n}=$$

$$\frac {(1/2)+(27)(3/4)^n}{(9)(3/4)^n+(4)}\to 1/8$$

Answer 2

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 $N/D=1/8$.

Answer 3

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} $