# Proof check: quotient rule for convergent sequences

I've found a few proofs online similar to mine leading me to believe mine is OK, but I'm not sure if there are any incorrect steps. Here's my attempt:

Prove $\dfrac{x_n}{y_n} \rightarrow \dfrac{x^*}{y^*}$ as $n \rightarrow \infty$ provided that $(y_n) \neq 0$ and $y^* \neq 0$.

Proof:

Assume $x_n \rightarrow x^*$ and $y_n \rightarrow y^*$ as $n \rightarrow \infty$. Also assume $(y_n) \neq 0$ and $y^* \neq 0$. First, let $\epsilon > 0$ be given. Then, $\exists \quad N_x \in \Bbb N : \left\lvert x_n-x^* \right\rvert < \epsilon \quad \forall \quad n \ge N_x$ and $\exists \quad N_y \in \Bbb N : \left\lvert y_n-y^* \right\rvert < \epsilon \quad \forall \quad n \ge N_y$.

\begin{align*} \left\lvert \frac{x_n}{y_n} - \frac{x^*}{y^*}\right\rvert &\le \left\lvert \frac{x_ny^*-x_ny_n}{y_ny^*} \right\rvert + \left\lvert \frac{x_ny_n-y_nx^*}{y_ny^*} \right\rvert\\ &= \left\lvert \frac{x_ny_n-x_ny^*}{y_ny^*} \right\rvert + \left\lvert \frac{x_ny_n-y_nx^*}{y_ny^*} \right\rvert\\ &= \left\lvert \frac{x_n}{y_ny^*} \right\rvert\left\lvert y_n-y^* \right\rvert + \left\lvert \frac{1}{y^*} \right\rvert\left\lvert x_n-x^* \right\rvert \qquad \qquad (i)\\ \end{align*}

We must find a lower bound for denominator of $\left\lvert \frac{x_n}{y_ny^*} \right\rvert$. Since $(y_n)$ is convergent, we know that there exists $N_* \in \Bbb N$ such that $\lvert y_n-y* \rvert < \frac{\lvert y^* \rvert}{2}$ for all $n \ge N_*$. By the reverse triangle inequality, $\lvert y^* \rvert - \lvert y_n \rvert < \frac{\lvert y^* \rvert}{2} \implies \lvert y_n \rvert > \frac{\lvert y^* \rvert}{2} \qquad \qquad (ii)$.

Now, substituting (ii) into (i) and using the fact that $(x_n)$ is bounded (because it's convergent) i.e. $\lvert x_n \rvert \le C$ for all $n \in \Bbb N$, and also the fact that $(x_n)$ and $(y_n)$ are convergent sequences, we get for $n \ge max\{N_x, N_y, N_*\}$:

$\left\lvert \frac{x_n}{y_n} - \frac{x^*}{y^*} \right\rvert < \frac{2C\epsilon}{\left\lvert y^* \right\rvert^2} + \frac{\epsilon}{\left\lvert y^* \right\rvert} = \frac{\epsilon}{\left\lvert y^* \right\rvert}(\frac{2C}{\left\lvert y^* \right\rvert}+1)$

The right hand side of this inequality gets arbitrarily small as epsilon becomes arbitrarily small, hence we can conclude that the original statement is indeed true. $\square$

Many thanks for any help.

• Explain what $x*,y*$ are please, – zhw. Dec 21 '15 at 22:22
• The limits of $(x_n)$ and $(y_n)$, respectively. – mathphys Dec 21 '15 at 22:22
• It's OK, but you could write it better. I don't understand which lower bound you found for $y_ny^\star$, for instance, and this is arguably the main point of this proof. – Giuseppe Negro Dec 21 '15 at 22:23
• You might want to acknowledge a little more carefully how you (implicitly assume and) use the fact that $y^*\not=0$. – John Dawkins Dec 21 '15 at 22:28
• Please edit and put all relevant hypotheses in. – zhw. Dec 21 '15 at 22:30