Proving certain properties of limits of sequences using the definition of limits

limitsreal-analysissequences-and-series

I have been trying to prove the following properties for limits of sequences:

  1. Let $b \neq 0$. If $\lim x_n = a$ and $\lim \frac{x_n}{y_n} = b$, then $\lim y_n = \frac{a}{b}$.
  2. Let $a \neq 0$. If $\lim x_n = a$ and $\lim x_ny_n = b$, then $\lim y_n = \frac{b}{a}$.

I know that these are straightforward to prove using other elemental properties of limits of sequences, but I want to prove them using the definition and I have been not able to do it.

So far I have, for (1):

Let $\epsilon > 0$. We know that:

i)$\exists n_1 \in \mathbb{N}: n >n_1 \implies |x_n – a| < \epsilon$

ii)$\exists n_2 \in \mathbb{N}: n >n_2 \implies |x_n/y_n – b| < \epsilon \implies |x_n – by_n| < |y_n|\epsilon$

Now let $n_0 = \max\{n_1, n_2\}$. Then, for $n > n_0$:

\begin{aligned}
|y_n – a/b| &= |1/b||by_n-a| \\
&= |1/b||by_n + x_n – x_n – a| \\
&\leq |1/b|(|x_n – by_n| + |x_n – a|) \\
&<|1/b|(|y_n| + 1)\epsilon
\end{aligned}

which doesn't work, since it depends on the value $|y_n|$, so what I need to prove is that $|y_n|$ is bounded above by an $M > 0$. I know that both $x_n$ and $x_n/y_n$ are bounded since they are convergent, but I can't prove the last step that $y_n$ is bounded. Thanks in advance!

Best Answer

You're definitely on the right track.

Because $x_n/y_n \to b$ and $b$ is nonzero, there is a positive integer $W$ so that whenever $n \geq W$ one has $|x_n/y_n| \geq |b|/2$.

(Please note, this is a step away from the "epsilon-N" definition of the assumed limit, but you get there by putting $|b|/2$ in for the "epsilon" in that definition and manipulating a few inequalities to see that $|x_n/y_n - b| < |b|/2$ implies $|x_n/y_n - 0| \geq |b|/2$. This is also clear from a picture of the number line. Anyway, you'll later want to toss $W$ into that $\max$ that you've got going.)

For all $n \geq W$ it follows that $|y_n| \leq 2 |x_n|/|b|$.

And if you know that $x_n$ that is bounded...

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