5
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ $y_n\neq 0$ does not grant $y\neq 0$, so $\frac{x}{y}$ might be meaningless. $\endgroup$
    – Jack D'Aurizio
    Sep 8, 2018 at 15:48
  • 1
    $\begingroup$ ... because of the condition (which you didn't mention) that $y\ne 0$, you can bound from above with a constant the quantity $\frac1{\lvert y_n\rvert}$... $\endgroup$
    – user562983
    Sep 8, 2018 at 15:48
  • 1
    $\begingroup$ @JackD'Aurizio It's just the way the question was written in the textbook. Yes, but you are right. $\endgroup$
    – OGC
    Sep 8, 2018 at 15:49

2 Answers 2

Reset to default
3
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ We need to be careful here, first of all, the signs, and second of all, $y$ might be $1$ $\endgroup$
    – Jakobian
    Sep 8, 2018 at 16:05
  • 2
    $\begingroup$ Suggestions: Instead of $y_n$, consider $|y_n|$. Choose $\epsilon = |y|/2$ $\endgroup$
    – Jakobian
    Sep 8, 2018 at 16:07
  • $\begingroup$ @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. $\endgroup$
    – Curious student
    Sep 8, 2018 at 16:22
  • $\begingroup$ Maybe you can use $\frac{}{}$ in order to make it more readable. Otherwise good answer, +1 $\endgroup$
    – EuklidAlexandria
    Sep 8, 2018 at 16:31
  • $\begingroup$ @ShubhamRJ Interesting way of proving. Answer accepted. $\endgroup$
    – OGC
    Sep 8, 2018 at 21:45
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – OGC
    Sep 8, 2018 at 21:46
  • $\begingroup$ @OGC Yes, you have to the product rule. In most courses the product rule is proved first and then the quotient rule. $\endgroup$
    – EuklidAlexandria
    Sep 8, 2018 at 22:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .