# Simple limit proof for normed components

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.

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.
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: $$\left|\|{\bf r}(t)\|-\|{\bf L}\|\right| \leq \|{\bf r}(t)-{\bf L}\|<\epsilon.$$