Suppose $\left \{ x_{n} \right \}$ and $\left \{ y_{n} \right \}$ converge to the limits $x$ and $y$, respectively. Also, suppose that $y_{n}$'s are nonzero. I want to show that the sequence $\left \{ \frac{x_{n}}{y_{n}} \right \} \rightarrow \frac{x}{y}$.
Let $\epsilon >0$. We can pick an $N$ such that $n \geq N$ $\Rightarrow \left | \frac{x}{y} – \frac{x_{n}}{y_{n}}\right | < \epsilon$.
I start from $$ \left | \frac{x}{y} – \frac{x_{n}}{y_{n}}\right |$$
$$=\left | \frac{x}{y} – \frac{x}{y_{n}} + \frac{x}{y_{n}} – \frac{x_{n}}{y_{n}}\right | $$
$$= \left | \frac{x\left (y_{n}- y \right )}{y y_{n}} + \frac{\left ( x-x_{n} \right )}{y_{n}}\right | $$
$$\leq \left | \frac{x}{y} \right | \frac{1}{\left | y_{n} \right |} \left | y-y_{n} \right | + \frac{1}{|y_{n}|}\left | x-x_{n} \right | $$ by triangle inequality.
I know that I want to make the two terms $\frac{\epsilon}{2}$ + $\frac{\epsilon}{2}$. Since $y_{n}$ converges to $y$, can make $\left | y – y_{n} \right |< \frac{\epsilon \left | y \right |}{\left ( \left | x \right | + 1\right )}$ for some $N_{1}$.
But the $y_{n}$ term is giving me a problem, which I can't get rid off.
Any suggestions on how to proceed from here (or approaching the problem from a different angle) would be greatly be appreciated.
Best Answer
As $y_n\neq 0 ,\forall n\in N$ also it is important to have condition that $y\neq 0$ as we example $y_n:=\frac{1}{n} $ which satifies above but $\frac{1}{y_n} $diverges.
SO $|y_n|>0$
As $y_n\to y $
so we have$ \forall \epsilon>0 ,\exists N_1>0 $such that $|y_n-y|<\epsilon $ for $n>N_1$
so after $N_1$ terms $|y|-\epsilon<|y_n|<|y|+\epsilon$
SO you have bound for $\frac{1}{y_n} <\frac{1}{|y-r|}$ for that N. Here $r=|y|/2$
SO $\frac{1}{|y_n|} <2/|y|$
Choose $|y-y_n|<\epsilon \frac{|y|^2}{|2x|}$ for $n>N_1$
Other term also follows same