proving $\lim(1/y_n)=1/y$

Question

I have a question on the proof of $\lim (1/y_n)=1/y$ under $\lim y_n=y$;

$|y_n-y| < \epsilon |M|/|y|$

$y_n$ is convergent so that it's bounded by some number $|M|$;

$|1/y_n-1/y|=\frac{|y_n-y|}{|y_n||y|}<\epsilon$

for some natural number.

Where is the wrong this proof?

I got a idea on the proof of $\lim(x_n y_n)=xy$.

Answer 1

It doesn't help that $|y_n|<M$ for all $n$. You need that $|y_n|>c$ for some $c>0$ and almost all $n$. You get this from $|y|\ne 0$ (a condition that is missing from the problem statement!)

Answer 2

You need $y\neq 0$, of course. Hence there is some $\eta>0$ and $N$ such that $|y_n| \ge \eta$ for all $n \ge N$ (hence $|y| \ge \eta$ as well).

Then we have $|{1 \over y_n} - {1 \over y}| = | { y-y_n \over y y_n } | \le {1 \over \delta^2} |y-y_n|$.

Now choose $N' \ge N$ such that $|y-y_n| < \delta^2 \epsilon$, whence we have $|{1 \over y_n} - {1 \over y}| < \epsilon$.

Answer 3

The standard proof is as follows: $$\frac{|y_n-y|}{|y_n||y|} \tag{1}$$

Let $|y_n-y|<|y|/2$ note that $y\not =0$ this is absolutely necessary. So $|y_n|>|y|/2$, which is the same as $1/|y_n|<2/|y|$, this is possible since $|y_n|>0$, and we get our desired upper bound. Putting this in (1) we have

$$\frac{|y_n-y|}{|y_n||y|}<\frac{2|y_n-y|}{y^2} \tag{2}$$

Now If $|y_n-y|<\min\left(|y|/2,y^2\varepsilon/2 \right)$. Thus

$$\frac{2|y_n-y|}{y^2}< \varepsilon$$