Let $(b_n)_{n=m}^\infty$ be a sequence of real numbers and let $L$ be a real number such that $\lim_{n\to\infty}b_n = L$. Let also $b_n \neq 0$ for all n and $L \neq 0$. Show that
\begin{equation*}
\lim_{n\to\infty}\frac{1}{b_n} = \frac{1}{L}.
\end{equation*}
I do not understand the hint behind the exercise. The hint states that one should prove that the sequence is bounded away from zero. However, if $b_n \neq 0$ for all $n$ why do I need to prove such a thing? In particular, can someone point out where the mistake is in my reasoning?
My proof:
Since $\lim_{n\to\infty} b_n = L$ the sequence is bounded; i.e., there exists a $B>0$ such that $|b_n|\leq B$ for all $n$. Observe that
\begin{align}
|\frac{1}{b_n} – \frac{1}{L}| &= |\frac{L-b_n}{b_nL}|\\
&= \frac{|b_n – L|}{|b_n||L|}\\
&\leq \frac{|b_n – L|}{B|L|}.
\end{align}
Choosing now an appropriate $\epsilon$ for $|b_n – L|$ like $\epsilon B|L|$ should be sufficient to conclude the reasoning.
Best Answer
Your last step goes from $|b_n| \leq B$ to $\frac{1}{|b_n|} \leq \frac{1}{B}$, which does not follow (the inequality goes the wrong way).
If you knew that $|b_n| \geq B$ for all large enough $n$ then you'd be able to use that estimate to get $\frac{1}{|b_n|} \leq \frac{1}{B}$ as you wished. Showing $|b_n| \geq B$ for all large enough $n$ is exactly what the hint you were given means.