$\lim x_n = a$, $\lim \frac{x_n}{y_n}=b$ then $\lim y_n = \frac{a}{b}$

I must prove:

$\lim x_n = a$, $\lim \frac{x_n}{y_n}=b$ then $\lim y_n = \frac{a}{b}$

Well, I know that

$$\lim x_n = a \implies |x_n-a|<\epsilon$$

$$\lim \frac{x_n}{y_n} = b \implies |\frac{x_n}{y_n}-b|<\epsilon$$

I need to use these inequalities to somehow arrive at:

$$|y_n-\frac{a}{b}|<\epsilon_0$$

But I cannot see any obvious algebraic relation between those absolute values. ANy helps?

• It was my mistake. It's right as you've written it. I've just had one too many glasses of milk. As a hint, consider what it means to take the ratio of those limits. From what you know, the sequence $x_n$ is Cauchy. You also know that the sequence $z_n = \frac {x_n} {y_n}$ is Cauchy. Could $z_n$ be Cauchy if $y_n$ weren't Cauchy? Now, you should be certain that $\lim y_n$ exists. You can also define the sequence $y_n$ in terms of the other two sequences $x_n, z_n$. You get $$y_n = \frac{z_n}{x_n}$$. From there, you should be able to show that $y_n$ converges to $\frac a b$ within $\epsilon_0$. – Axoren Apr 28 '16 at 5:18
• @Axoren I cannot use any Cauchy hypotesis – user335283 Apr 28 '16 at 5:19
• That is unfortunate. That's the easiest way to show it. One thing you could do is replace the absolute values with the square-root of a square. $|x_n - a| = \sqrt{(x_n - a)^2}$. This enforces positivity but is much more algebraically workable. – Axoren Apr 28 '16 at 5:22

HINT:

\begin{align} \left|y_n -\frac ab\right|&=\left|\frac{x_n}{\frac{x_n}{y_n}} -\frac ab\right|\\\\ &=\left|\frac{b(x_n-a)-a\left(\frac{x_n}{y_n}-b\right)}{b\frac{x_n}{y_n}}\right| \end{align}

Now, apply the triangle inequality.

$b$ is not be 0. $a$ and $b$ are real numbers.

$x_n=a+\alpha$, where $\alpha \to 0$, when $n \to \infty$.

$y_n=\frac{x_n}{b+\beta}$, where $\beta \to 0$, when $n \to \infty$.

$y_n=\frac{a+\alpha}{b+\beta}$. $y_n \to \frac{a}{b}$.