$\lim \frac{U_n}{V_n} = 1$, How to prove that $U_n$ and $V_n$ have the same sign for a $n> N$

Question

With $V_n \neq 0 $,

We have for an $n> N$: $|\frac{U_n}{V_n} - 1| < \epsilon $

and: $ 1 - \epsilon < \frac{U_n}{V_n} < 1 + \epsilon $

If we'll take $\epsilon = \frac{1}{2}$, we'll have:$ \frac{1}{2} < \frac{U_n}{V_n} < \frac{3}{2} $

Which means:$ \frac{U_n}{V_n} > 0$

Is this correct? Thank you.

Answer 1

I think it's better the following.

Given that for all $\epsilon>0$ there is $N$ for which for all $n>N$ we have: $$1-\epsilon<\frac{U_n}{V_n}<1+\epsilon$$

Let $\epsilon=\frac{1}{2}$.

Hence, there is $N$, for which for all $n>N$ we have $$1-\frac{1}{2}<\frac{U_n}{V_n}<1+\frac{1}{2}$$ and we are done!