I've been tasked with proving the above question and I was hoping some one could get some feedback. I will confess this is a proof for a similar question I've attempted to convert into the above so it will likely be a bit weird.
$x_n \rightarrow x$, means for any $\epsilon _1 > 0$then there is an $N_{\epsilon_1}$ so that $n > N_1 \implies |x_n-x| <\epsilon_1$. And $x_ny_n \rightarrow z$ means for any $\epsilon_2 >0$ then there is an $N_{\epsilon_2}$ so that $n>N_2 \implies |x_n y_n – z|$. So if $n>max(N_{\epsilon_1},N_{\epsilon_2})$ then $|x_n y_n – z|<\epsilon_1 \epsilon_2$.
And meanwhile
$\left| y_n – \frac{z}{x} \right|=|x_n y_n – z| – |x_n – x| \leq |x_n y_n – z| +|x_n – x| $
So for any any $\epsilon > 0$ let $\epsilon_1=\epsilon_2 = \epsilon$ and let $N_{\epsilon_1}$ and $N_{\epsilon_2}$ be as above and $N = max(N_{\epsilon_1},N_{\epsilon_1})$.
Then
$n>N \implies \left|y_n – \left(\frac{z}{x}\right)\right| = ||x_n y_n – z| – |x_n – x|| \leq |x_n y_n – z| + |x_n – x| < \epsilon_1 \epsilon_2 = \epsilon \epsilon=\epsilon $
Therefore $y_n \rightarrow \frac{z}{x}$.
Any tips or insights would be much appreciated! As I am rather new to all this.
Best Answer
We first prove $(y_n)$ is bounded. Since $(x_n)$ converges to a nonzero value $x$, there is an $N_x$ such that for all $n>N_x$, we have $|x_n|>M_x>0$ for some $M_x$. Now since $(x_ny_n)$ converges to $z$, by definition, there is an $N>N_x$ such that for all $n>N$ we have $|x_ny_n -z| <4$. Surely, by the reverse triangle inequality, we have $|x_ny_n|<4+|z|$, and so $M_x|y_n|<|x_n||y_n|<4+|z|$, which then implies $|y_n|<M_x^{-1}(4 + |z|)$. Hence if we let $M=\max\{|y_1|,|y_2|,\dots,|y_N|,M_x^{-1}(4 + |z|)\}$, we can clearly see the sequence $(y_n)$ is bounded by this $M$.
Now we can easily prove $(y_n)$ converges, and yes, your intuition is correct in thinking the limit will be $zx^{-1}$. Let $M>0$ bound $(y_n)$ (which exists as above). Let $\epsilon>0$ be given. Then there exist $N_1,N_2$ such that for $n>N_1$ we have $|x_n-x|<\dfrac{\epsilon |x|}{2M}$ and for $n>N_2$ we have $|x_ny_n-z|<\dfrac{\epsilon |x|}{2}$. Then for $n>\max \{N_1,N_2\}$ we have: \begin{align} |y_n-zx^{-1}|&=|x|^{-1}|xy_n-z| \\ &= |x|^{-1}|xy_n-z+x_ny_n-x_ny_n| \\ &= |x|^{-1}|xy_n-x_ny_n+x_ny_n-z| \\ &\leq |x|^{-1}(|xy_n-x_ny_n|+|x_ny_n-z|) \\ &= |x|^{-1}(|y_n||x-x_n|+|x_ny_n-z|) \\ &< |x|^{-1}\left(|y_n|\frac{\epsilon |x|}{2M}+\frac{\epsilon |x|}{2}\right) \\ &< |x|^{-1}\left(M\frac{\epsilon |x|}{2M}+\frac{\epsilon |x|}{2}\right) = \epsilon \end{align}