determine the limit of sequence?

Question

Need help on how to determine the limits, if it exists for this sequence. Have no idea where to start. Thanks in advance smart people!

$$A_n = \dfrac {(n+2)^{2n}}{(n^2-n-6)^n}$$

Answer 1

$$a_n=\frac{(n+2)^{2n}}{(n^2-n-6)^n}=\frac{(1+\frac 4 n+\frac 4{n^2})^{n}}{(1-\frac1n-\frac6{n^2})^n}$$ so $$\log(a_n)=n\left(\log\left(1+\frac 4 n+\frac 4{n^2}\right)-\log\left(1-\frac1n-\frac6{n^2}\right)\right)\sim_\infty n\left(\frac 4 n+\frac 1n\right)=5$$ so $$\lim_{n\to\infty}a_n=e^5$$

Answer 2

hint: $$A_n=\frac{(n+2)^{2n}}{(n^2-n-6)^n}=\left(\frac{(n+2)(n+2)}{(n+2)(n-3)}\right)^n$$ $$=\left(\frac{n+2}{n-3}\right)^n=\left(\frac{1+2/n}{1-3/n}\right)^n$$

Answer 3

Write $A_n = \left( (n+2)^2/(n^2-n-6) \right)^n = \left( (n+2)/(n-3) \right)^n = \left( (n-3+5)/(n-3)\right)^n = \left(1 + 5/(n-3)\right)^n$.

This limit should look familiar.