# Proving that the sequence $\left \{ \frac{x_{n}}{y_{n}} \right \} \rightarrow \frac{x}{y}$

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.

• $y_n\neq 0$ does not grant $y\neq 0$, so $\frac{x}{y}$ might be meaningless. – Jack D'Aurizio Sep 8 '18 at 15:48
• ... because of the condition (which you didn't mention) that $y\ne 0$, you can bound from above with a constant the quantity $\frac1{\lvert y_n\rvert}$... – Saucy O'Path Sep 8 '18 at 15:48
• @JackD'Aurizio It's just the way the question was written in the textbook. Yes, but you are right. – OGC Sep 8 '18 at 15:49

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

• We need to be careful here, first of all, the signs, and second of all, $y$ might be $1$ – Jakobian Sep 8 '18 at 16:05
• Suggestions: Instead of $y_n$, consider $|y_n|$. Choose $\epsilon = |y|/2$ – Jakobian Sep 8 '18 at 16:07
• @Rumpelstiltskin Thanks A lot.A got New Insights for writting proof.Actually Now I am feeling for Learning Mathematics This best place As here by writing solution Error get encountered. – MathLover Sep 8 '18 at 16:22
• Maybe you can use $\frac{}{}$ in order to make it more readable. Otherwise good answer, +1 – EuklidAlexandria Sep 8 '18 at 16:31
• @ShubhamRJ Interesting way of proving. Answer accepted. – OGC Sep 8 '18 at 21:45

You can proof $$\lim_{n\to\infty}\frac{1}{y_n}=\frac{1}{y}$$ and then use the product rule.

• Yes, this seems to be the standard way of proving this. Might also have to prove that $\left \{ x_{n} y_{n}\right \} \rightarrow xy$ before applying it. – OGC Sep 8 '18 at 21:46
• @OGC Yes, you have to the product rule. In most courses the product rule is proved first and then the quotient rule. – EuklidAlexandria Sep 8 '18 at 22:58