Tao Analysis 1 Exercise 6.1.8. 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.
 A: 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.
A: $|b_n|\le M$ implies $\frac{1}{|b_n|}\ge \frac{1}{M}$
$\frac{|b_n - L|}{|b_n||L|}
\leq \frac{|b_n - L|}{B|L|}$ ❌
Hint for the correct proof:
$b_n\to L$ and $L\neq 0$ implies $\exists N\in\Bbb{N}$ such that $\forall n\ge N$ ,$|b_n|\ge \frac{|L|}{2}$
( use $\epsilon=\frac{|L|}{2}$ in $b_n\to L$ )
A: The bound you have imposed is that $\lvert b_n\rvert\leq B$ for all $n$, and while this can be done, it's not what you want, and is not what the hint is telling you. Indeed it is not true that
$$\frac{\lvert b_n-L\rvert}{\lvert b_n\rvert\lvert L\rvert}\leq\frac{\lvert b_n-L\rvert}{B\lvert L\rvert},$$
but the reverse inequality is true. That is why it does not help you. Now what the hint tells you to do is show the existence of some $C>0$ such that $C\leq \lvert b_n\rvert$ for all $n$. If you can do this, then just replace $B$ with $C$ in your argument and all is good.
