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if $\mathbf r(t)$ is a vector valued function satisfying $\lim_{t\to c} \mathbf {r}(t) = \mathbf L$, then prove that $\lim_{t\to c} ||\mathbf r(t)|| = ||\mathbf L||$

Hint: the triangle inequality is relevant here

Attempt:

All I could think of is the delta epsilon definition of what it means for a limit to exist...actually thinking some more about it, would I use this variant of the triangle inequality:

$| |x| - |y| | \le |x - y|$ which in my case would mean since $\lim_{t\to c} \mathbf {r}(t) = \mathbf L$, already exists, it bounds the normed version.

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You have $\|x\| \le \|x-y\|+\|y\|$ and so $\|x\|-\|y\| \le \|x-y\|$. Switching the roles of $x,y$ gives $|\|x\|-\|y\| | \le \|x-y\|$, so we see that the norm is continuous.

Since $|\|r(t)\|-\|L\| | \le \|r(t)-L\|$, you obtain the desired result.

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You're on the right way. Let $\epsilon > 0$. There exists $\delta > 0$ such that $|t-c|<\delta$ implies $\|{\bf r}(t)-{\bf L}\|<\epsilon$. If $|t-c|<\delta$, then: $$\left|\|{\bf r}(t)\|-\|{\bf L}\|\right| \leq \|{\bf r}(t)-{\bf L}\|<\epsilon.$$

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  • $\begingroup$ I apologize i just got really sloppy with the notation. I'm typing from a tablet, computer died on me, so the tablet isn't the most latex friendly machine. $\endgroup$ – dc3rd Jun 4 '15 at 18:51

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