Calculating limit using sub sequences $\lim\limits_{n\to\infty }\frac{(-5)^{n}+2\cdot(-2)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$

Find the following limit if it exists , if it does not exist (even not infinity) explain if it does exist then also find all the Sub sequential limits. $\lim\limits_{n\to\infty }\frac{(-5)^{n}+2\cdot(-2)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$

Before I show what I tried , Sorry if some words and terms are not translated right as I could not find the terms after searching , would appreciate if someone can help edit if there are mistakes.

My try:

First let $a_n=\frac{(-5)^{n}+2\cdot(-2)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$ then I wrote it in a more "comfortable" way so $\frac{5^n\cdot(-1)^{-n}+(2)^{n+1} \cdot(-1)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$ and then $\frac{(-1)^{n}\cdot(5^n+(2)^{n+1})+3}{5^{n+1}+2\cdot(-3)^{n}+3}$

if we look at the subsequences $(a_{2n})$ and $(a_{2n+1})$ then they cover the sequence $(a_n)$ and if two sub sequences converge to the same limit and they cover the sequence $(a_n)$ then $(a_n)$ also converges.

1. $(a_{2n})$ = $\frac{(5^n+(2)^{2n+1})+3}{5^{2n+1}+2\cdot(3)^{2n}+3}$ so $\lim\limits_{n\to\infty }\frac{(5^n+(2)^{2n+1})+3}{5^{2n+1}+2\cdot(3)^{2n}+3}$ and using basic arithmetic limit rules $\lim\limits_{n\to\infty }\frac{5^{2n}}{5^{2n+1}}\cdot\frac{(1+\frac{2^{2n+1}}{5^{2n}}+\frac{3}{5^{2n}})}{(1+\frac{2\cdot(3)^{2n}}{5^{2n+1}}+\frac{3}{5^{2n+1}})}$ = $\frac{1}{5}$
2. $(a_{2n+1})$ = $\frac{(-1)^{2n+1}\cdot(5^{2n+1}+(2)^{2n+2})+3}{5^{2n+2}+2\cdot(-3)^{2n+1}+3}$ so we get $(a_{2n+1})$ = $\frac{-5^{2n+1}-(2)^{2n+2}+3}{5^{2n+2}-2\cdot(3)^{2n+1}+3}$ then to calculate the limit I also used basic arithmetic rules $\lim\limits_{n\to\infty}\frac{-5^{2n+1}-(2)^{2n+2}+3}{5^{2n+2}-2\cdot(3)^{2n+1}+3}$

$\lim\limits_{n\to\infty}\frac{5^{2n+1}}{5^{2n+2}}\cdot\frac{(-1-\frac{2^{2n+2}}{5^{2n+1}}+\frac{3}{5^{2n+1}})}{(1+\frac{2\cdot(3)^{2n+1}}{5^{2n+2}}+\frac{3}{5^{2n+2}})} = -\frac {1}{5}$

so I got that the limits of the subsequences are different , can I conclude from that that the limit of $(a_n)$ does not exist? is there a different way to approach this?

There's a theorem that you may or may not have proved yet that goes something like the following:

If $(a_n)$ converges to $\ell$, then any subsequence $(a_{n_k})$ converges to $\ell$ as well.

So, if there are two subsequences that converge to two different values, that must mean that the original sequence does not converge.

Double check your class notes and the textbook to see if you've already proved this (or another similar theorem), but from there you've already done the work to prove that it does not converge.