Epsilon delta proof (exam type question)

Question

I would like to prove the statement

Let $(x_n)$ and $(y_n)$ be convergent sequence with limit $x$ and $y$ respectively. If $x_n >1$ for all $n$ and $\lim(x_n y_n) = z$, then $$\lim(y_n) = \frac{z}{x}$$

My attempt: I observed that $$\vert y_n - \frac{z}{x} \vert = \vert \frac{xy_n-z}{x} \vert = \frac{1}{|x|}\vert xy_n-z \vert$$

From the assumption, I believe that I should get $|x_n y_n -z|$ in the expansion above. But I can't made it

Answer 1

Hint:

First since $x_n>1,\forall n$, we hanve $x\ge 1$.

Following your proof, you have

$$\frac{1}{|x|}| xy_n-z |=\frac{1}{|x|}| xy_n-x_ny_n+x_ny_n-z |\le \frac{1}{|x|}(| x-x_n||y_n|+|x_ny_n-z |)$$ Also note since $y_n$ is convergent, it's bounded. Then use convergence of $x_n$, $(x_ny_n)$

Note $y_n=\frac{x_ny_n}{x_n}$, in general you can show :

Given $\lim a_n=a,\lim b_n=b\not =0$, $b_n\not=0,\forall n$, we have $\lim \frac{a_n}{b_n}=\frac{a}{b}$.

Answer 2

Suppose $\lim \limits_{n \to \infty} x_{n}y_{n} = z$ with $x_{n} \to x$ and $y_{n} \to y$. We want to show $y_{n} \to \frac{z}{x}$.

First of all, $x_{n} \neq 0$ for all $n$ (why?). So we can divide by any $x_{n}$, so we have:

$|y_{n} - \frac{z}{x}| = |y_{n} - \frac{z}{x_{n}} + \frac{z}{x_{n}} - \frac{z}{x}| \leq |y_{n} - \frac{z}{x_{n}}| + |\frac{z}{x_{n}} - \frac{z}{x}|$.

$|y_{n} - \frac{z}{x_{n}}| = |\frac{y_{n}x_{n} - z}{x_{n}}|= \frac{1}{|x_{n}|}|y_{n}x_{n} - z| \leq |y_{n}x_{n} - z|$ Since $|x_{n}| >1$ for all $n$.

So, putting all of the work together gives $$|y_{n} - \frac{z}{x}| \leq |y_{n}x_{n} - z| + |\frac{z}{x_{n}} - \frac{z}{x}|$$

For the second term, we have $|\frac{z}{x_{n}} - \frac{z}{x}|=|\frac{zx - x_{n}z}{x_{n}x}| = |\frac{1}{x_{n}x}||zx - x_{n}z| \leq 1 \cdot |zx - x_{n}z| \leq |z||x - x_{n}|$ (since $x_{n} > 1$ for all $n$ implies $x >1$ (why?).

So, finally, putting everything together gives

$$|y_{n} - \frac{z}{x}| \leq |y_{n}x_{n} - z| + |\frac{z}{x_{n}} - \frac{z}{x}| \leq |y_{n}x_{n} - z| + |z||x - x_{n}|$$

So, given $\epsilon > 0$, choose $N$ so that the first term on the RHS is $< \frac{\epsilon}{2}$, the second term is $< \frac{\epsilon}{2|z|}$ (why can you do this?).

Answer 3

Note that $\large y_n = \large\frac{x_ny_n}{x_n}$ for all $n$ as ${x_n} >1$ $\forall n$.

Thus,

$\large|y_n - \frac{z}{x}|$ = $\large|\frac{x_ny_n}{x_n} - \frac{z}{x}| = |\frac{x_ny_nx -zx_n}{xx_n}|$ = $|\large \frac{x_ny_nx - zx_n - xz + xz}{xx_n}| \leq |\frac{x_ny_n- z}{x_n}| + |\frac{z}{x}||\frac{x-x_n}{x_n}|$

For $n$ large enough, we can make $\large |\frac{1}{x_n}|\leq |\frac{2}{x}|$

We can prove this as follows:

$|x| = |x - x_n + x_n| \leq |x-x_n| + |x_n|$. For $n$ sufficiently large, we can make $|x-x_n| < |x|$, so that $|x| = |x-x_n+x_n|\leq |x-x_n| + |x_n| \leq 2|x_n|$

So for $n$ sufficiently large $|x| \leq 2|x_n|$ so $\frac{1}{|x_n|} \leq \frac{2}{|x|}$

Let $\frac{2}{|x|} = c>0$ as ${x_n} >1$ $\forall n$

Then

$\large |y_n - \frac{z}{x}|$ = $\large|\frac{x_ny_n}{x_n} - \frac{z}{x}| = |\frac{x_ny_nx -zx_n}{xx_n}|$ = $|\large \frac{x_ny_nx - zx_n - xz + xz}{xx_n}| \leq |\frac{x_ny_n- z}{x_n}| + |\frac{z}{x}||\frac{x-x_n}{x_n}|$$\leq c(|x_ny_n -z| + |z||x_n-x|)$

But as $n\to\infty$ ${|x_ny_n - z|}\to0$ and ${|x_n-x|}\to0$

Thus, it is proved.

NOT PART OF ANSWER:

user46944 (why I disagree):

I still disagree. Your argument will still work if $|z| = 0$, this is what you had before: $|y_{n} - \frac{z}{x}| \leq |y_{n}x_{n} - z_{n}| + |\frac{z_{n}}{x_{n}} - \frac{z}{x}| \leq |y_{n}x_{n} - z_{n}| + |x||z_{n} - z| + |z||x_{n} - x| $

implying $|y_{n} - \frac{z}{x}| \leq |y_{n}x_{n} - z_{n}| + |\frac{z_{n}}{x_{n}} - \frac{z}{x}| \leq |y_{n}x_{n} - z_{n}| + |x||z_{n} - z| + |z||x_{n} - x| \leq |y_nx_n - z| + (|x|+1)(|z-z_n|) + |z||x_n -x|$

Now if $|z| = 0$ the RHS becomes $|y_nx_n - z| + (|x|+1)(|z-z_n|)$ which can still be made arbitrarily small given $n$ sufficiently large.