HW 3

1. Scratch work We need to show that for any $M>0$ there exists some $N$ such that $n>N⟹n^2-5n>M$. Lets start off with the ending and work backwards. $n^2-5n>M$. Notice that $n<n^2-5n$ for $n>5$. Then it is sufficent to prove that $n>M$. This is true for $N=\max 5,M$ Proof:

1. Let $M>0$ and $N=\max \{5,M\}$
2. Then, for any $n>N$ we have $n>M$ and $n>6$
3. Notice, $n^2-5n>n>M$
4. Thus, for any $M>0$ we have an $N=\max \{5,M\}$ such that $n>N⟹n^2-5n>M$ 2. finished on ipad

3. #todo 4. finished on ipad 5. im gonna skip because fuck induction.