Question

(a)n=[(n-1)/(n^2+1)], calculate the 6 values of limbs, also Lim(o-infiniti)(an)

Answer 1

\[a_n=\frac{n-1}{n^2+1}\]?

Answer 2

ja

Answer 3

then
\[\lim_{n\to \infty}a_n=0\]since the degree of the numerator is less than the degree of the denominator
not sure what limbs are

Answer 4

ok

Answer 5

LIMBS IS chain

Answer 6

Does it mean to
"calculate \( a_0\), \( a_1\), \( a_2\), ...\( a_5\), and \( Lim_0^\infty a_n \) " ?

Answer 7

ja thats it

Answer 8

For n=0, a0=(0-1)/(0+1)=-1
n=1, a1=(1-1)/(1+1)=0
....
until n=5, a5=(5-1)/(5^2+1)=4/26=2/13