(1 pt) let an=n+1n+3. find the smallest number m such that: (a) |an−1|≤0.001 for n≥m m= 1997 (b) |an−1|≤0.1 for n≥m m= 23 (c) now use the limit definition to prove that limn→[infinity]an=1. that is, find the smallest value of m (in terms of t) such that |an−1|m. (note that we are using t instead of ϵ in the definition in order to allow you to enter your answer more easily). m= 2/t - 3t (enter your answer as a function of t)