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If $\langle X_n \rangle$ is a sequence in $\mathbb R^n$ then $\lim\limits_{n\to \infty}X_n=A$. Show that $\lim\limits_{n\to \infty} \| X_n \| = \| A \|$ .

My thought process, $\| X_i \|= (\sum_{i=1}^n x_i^2)^{(1/2)}$ therefore be the multiplicative property of limits $x_i^2 = a_i^2$. Therefore the limit of norm of the components is equivalent to the limit of the corresponding limit of each component of the sequence.

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your thought process is correct. The proof relies on the fact that element wise convergence iff convergence and the fact that norm is a continuous function of the components. – mathemagician Jan 16 '13 at 3:10
up vote 4 down vote accepted

If $x_n \to a$, then by definition $\|x_n - a\| \to 0$. But we have the subtriangle inequality, so for every $n$ $|\|x_n \| - \|a\| | \le \|x_n - a\|$. The result now follows by the squeeze theorem.

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I've always seen this referred to as the reverse triangle inequality. In any event, +1. – Benjamin Dickman Jan 16 '13 at 3:08
Ahhh,so clever. – user5208 Jan 16 '13 at 3:12

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