I'm having trouble with this particular proof regarding sequences:
I'm assuming the result from (a) leads to (b) which eventually leads us to what we want to end up proving, (c).
I've been playing around with the definition of a sequence in part (a), but not exactly sure what to do with it when it comes to the delta value.
Any help/hints/advice would be much appreciated.
Thanks!
We know that $\exists N \in \mathbb{N} \text{ st } \{n \geq N \Longrightarrow |b_n-M|< \frac{|M|}{2}\}.$ We can prove that this implies $|b_n|>\frac{|M|}{2}$
Prove by contradiction. Suppose $\exists n \geq N$ such that $|b_n|\leq \frac{|M|}{2}.$ Then by the above inequality, $\left|\frac{|M|}{2}-M\right|<\frac{|M|}{2},$ but this is a contradiction. Thus we have proven that $\exists N \in \mathbb{N} \text{ st } \{n \geq N \Longrightarrow |b_n|>\frac{|M|}{2}\}.$
But we need a lower bound for all the numbers in the sequence. But that's not hard to do. Let $$\delta := \min(|b_1|,|b_2|,...,|b_{N-1}|,\frac{|M|}{2}).$$ Then certainly $|b_n| \geq \delta$ for all $n$ so we have shown (a).
(b) is easy so I won't do it here.
For (c), we use the definition of limits once again. We know that given $\epsilon >0,$ $\exists N' \in \mathbb{N} \text{ st } \{n \geq N' \Longrightarrow |b_n-M|< \epsilon\}$ because $b_n \rightarrow M.$ So then, $\forall n>N',$ $$\left\lvert\frac{1}{b_n}-\frac{1}{M}\right\rvert=\frac{|M-b_n|}{|b_nM|}<\frac{2\epsilon}{\delta|M|}$$ by (b). $\frac{2}{\delta|M|}$ is just a constant, so we're done.
