Write out the first four terms of the following sequences, then evaluate the limits.

1. $a _ { n } = \frac { 2 - n } { 3 + n }$. Then find $\lim\limits _ { n \rightarrow \infty } a _ { n }$.

   Steps (a):

   $a_1=\frac{2-1}{3+1}=?$

   $a_2=\frac{2-2}{3+2}=?$

   $a_3=\frac{2-3}{3+3}=?$

   $a_4=\frac{2-4}{3+4}=?$

   More Help (a):

   1. Split the fraction into two fractions:

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

   2. Use the dominant term idea to find the limit.

2. $a _ { n } = \frac { ( - 2 ) ^ { n } } { 3 ^ { n } }$. Then find $\lim\limits_ { n \rightarrow \infty } a _ { n }$.

   Steps (b):

   $a_1=\frac{(-2)^1}{3^1}=?$

   $a_2=\frac{(-2)^2}{3^2}=?$

   $a_3=\frac{(-2)^3}{3^3}=?$

   $a_4=\frac{(-2)^4}{3^4}=?$

   Hint (b):

   The dominant term is in the denominator.

3. $a _ { n } = \frac { ( - 2 ) ^ { n } } { 2 ^ { n } }$. Then find $\lim\limits _ { n \rightarrow \infty } a _ { n }$.

   Steps (c):

   $a_1=\frac{(-2)^1}{2^1}=?$

   $a_2=\frac{(-2)^2}{2^2}=?$

   $a_3=\frac{(-2)^3}{2^3}=?$

   $a_4=\frac{(-2)^4}{2^4}=?$

   Hint (c):

   The dominant term is in both the denominator and the numerator.