Can you find the limit of the following sequences

a_{n = n^3 e^-2n
an = 2^2n/3^n}

e^-2n=1/e^2n and e^2n=1+2n+(2n)^2/2!+(2n)^3/3!+(2n)^4/4!+…
so e^2n>2n^4/3 and n^3 e^-2n < n^3/(2n^4/3)=3/(2n) and this ->0
as n-> inf.

2^2n=4^n so a(n)=4^n/3^n=(4/3)^2 and since (4/3)>1 a(n)-> infinity
as n-> infinity.

one way is if u plug in values of n to see whats happening
1. n^3 e^-2n
n=1: e^-2
n=2: 8e^-4
n=3: 27e^-6
u can even graph it to see the graph, the LImit goes to zero

same way
n=1; 4/3
n=2; 16/9
n=3; 64/27
and if u graph it, its increasing so the Limit goes to Infinity