Suppose $\lim x_{n}=x$ and $x\ne0$. Show that there is an $N\in\mathbb{N}$ so that if $n\ge N$ then $|x_{n}|\ge\frac{|x|}{2}.$

Question

Suppose $\lim x_{n}=x$ and $x\ne0$. Show that there is an $N\in\mathbb{N}$ so that if $n\ge N$ then $|x_{n}|\ge\frac{|x|}{2}.$

Since $\lim x_{n}=x$ there exists an $N \in \mathbb{N}$ so that if $n \geq N$ then $|x_n-x|< \frac{|x|}{2}$. So when $n \geq N$ we have that

$\frac{|x|}{2}>|x_n-x|| \geq ||x_n|-|x|| \geq |x_n|-|x|$

Thus we have that $\frac{|x|}{2} \geq |x|-|x_n|$ so we have taht $|x_n| \geq |x|- \frac{|x|}{2}$ therefore $|x_n| \geq \frac{|x|}{2}$

Answer 1

It's almost fine, there is a small typo @JoséCarlosSantos pointed at, but hopefully the version below is cleaner.

If $\lim\limits_{n\rightarrow\infty}x_n=x$ then $\lim\limits_{n\rightarrow\infty}|x_n|=|x|$, it is easy to see this from $$0\leq||x_n|-|x||\leq |x_n-x|<\varepsilon$$

So we can say $$\forall \varepsilon >0, \exists N(\varepsilon): ||x_n|-|x|| < \varepsilon, \forall n>N(\varepsilon)$$ which is $$|x|-\varepsilon < |x_n| < |x|+\varepsilon$$ Now, take $\varepsilon =\frac{|x|}{2}>0$