(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|≤0.00001 for n≥M M= 200003 (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)

We want to prove[tex]\displaystyle\lim_{n\to\infty}\frac{n+1}{n+3}=1[/tex]which is akin to saying that, for any given [tex]\varepsilon>0[/tex], we guarantee that[tex]\left|\dfrac{n+1}{n+3}-1\right|<\varepsilon[/tex]for all [tex]n[/tex] exceeding some threshold [tex]N=N(\varepsilon)[/tex].We want to end up with[tex]\left|\dfrac{n+1}{n+3}-1\right|=\left|-\dfrac2{n+3}\right|=\dfrac2{n+3}<\varepsilon[/tex][tex]\implies\dfrac{n+3}2>\dfrac1\varepsilon\implies n>\dfrac2\varepsilon-3[/tex]which suggests that we guarantee that [tex]a_n[/tex] is arbitrarily close to 1 if [tex]N=\left\lceil\dfrac2\varepsilon-3\right\rceil[/tex]. (Take the ceiling to ensure [tex]N[/tex] is a natural number.)Now for the proof: Let [tex]\varepsilon>0[/tex] be given. Then for [tex]n>N=\left\lceil\dfrac2\varepsilon-3\right\rceil[/tex] we have[tex]n>\left\lceil\dfrac2\varepsilon-3\right\rceil\le\dfrac2\varepsilon-3\implies\dfrac2{n+3}<\varepsilon\implies\left|\dfrac{n+1}{n+3}-1\right|<\varepsilon[/tex]QED