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 $.


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.

  • $\begingroup$ Explain what $x*,y*$ are please, $\endgroup$ – zhw. Dec 21 '15 at 22:22
  • $\begingroup$ The limits of $ (x_n) $ and $ (y_n) $, respectively. $\endgroup$ – mathphys Dec 21 '15 at 22:22
  • $\begingroup$ 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. $\endgroup$ – Giuseppe Negro Dec 21 '15 at 22:23
  • $\begingroup$ You might want to acknowledge a little more carefully how you (implicitly assume and) use the fact that $y^*\not=0$. $\endgroup$ – John Dawkins Dec 21 '15 at 22:28
  • $\begingroup$ Please edit and put all relevant hypotheses in. $\endgroup$ – zhw. Dec 21 '15 at 22:30

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