Prove that if $\displaystyle x_n \to x $ and $\displaystyle x>0 $ , then there is $\displaystyle k \in N $ such that $\displaystyle \frac {x}{2} < x_n < \frac{3x}{2}, \forall n \geq k $
By definition, since $\displaystyle x_n$ converges to $\displaystyle x$, we can find $\displaystyle k$ such that $\displaystyle |x_n-x|<\frac{x}{2}$ whenever $\displaystyle n\geq k$, i.e.
$\displaystyle \frac{-x}{2}<x_n-x<\frac{x}{2}$
whenever $\displaystyle n\geq k$. Adding $\displaystyle x$ throughout this inequality we get
$\displaystyle \frac{x}{2}<x_n<\frac{3x}{2}$
whenever $\displaystyle n\geq k$.