Suppose a seqence $\{x_n\}$ converges. Prove that $\lim_{n→∞}|x_n| = |\lim_{n→∞} x_n|$ 
Suppose a seqence $\{x_n\}$ converges. Prove that $\lim_{n→∞}|x_n| = |\lim_{n→∞} x_n|$

I have been trying this proof by playing around with the definition of a limit and using the triangle inequality, but have not been able to figure it out.
 A: The absolute value is continuous on $\mathbb{R}$, so the equality holds. Alternatively, by the reversed triangle inequality we have $$\left| \left| x_n\right| - \left| \lim_{n \to \infty}x_n\right|\right| \leqslant \left| x_n - \lim_{n \to \infty}x_n\right| < \varepsilon$$ for $n > N$, where $N \in \mathbb{N}$ is choosen so, such that for fixed $\varepsilon > 0$ $$\left| x_n - \lim_{n \to \infty}x_n\right|< \varepsilon$$ This is possible, since $x_n$ is convergent.
A: Suppose $\{x_n\}_n$ converges t0 $x$ so $\lim_{n\rightarrow \infty} x_n = x$. Then $\big |\lim_{n\rightarrow \infty} x_n \big | = |x|$ so we need to prove that $\lim_{n\rightarrow \infty} |x_n|=|x|$. 
By the triangle inequality for any real numbers $a$ and $b$ we have $ |a| \le |a-b| + |b| \Rightarrow |a|-|b| \le |a-b| $ similarly $|b| \le |a-b| +|a| \Rightarrow |b|-|a| \le |a-b|$ and from these two inequalities we can conclude that $||a|-|b|| \le |a-b|$
So since $x_n\rightarrow x $ then for any $\epsilon > 0$ there exists an $N\in \Bbb N$ so for $n\ge N$ we have $ ||x_n|-|x|| \le |x_n-x| \le \epsilon $ which proves $\lim_{n\rightarrow \infty} |x_n|=|x|$
