lim r(t) = L implies lim ||r(t)|| = ||L||

Question

The question is:

if $\lim_{t\rightarrow t_0}\vec r(t) = \vec L$ show that $\lim_{t\rightarrow t_0} \|\vec r(t)\| = \|\vec L\|$

So here's where I am so far:

let $\vec r(t) = (f_1, f_2,...f_n)$ be such that the limit at a point $t_0$ is $\vec L$. Then by definition for all $\epsilon >0$ there exists $\delta>0$ such that $$\|\vec r(t) - \vec L\|<\epsilon$$ whenever $|t-t_0|<\delta$

I know I'm not very far but I don't know how to proceed, my initial thought was to mimic the proof of showing convergence implies absolute convergence, but don't seem to be getting anywhere. Please give only hints and no complete proofs as this is homework and i don't want to cheat. Thanks

Answer 1

By definition $\lim\limits_{t \to t_0} \vec{r}(t) = \vec{L} \Rightarrow$ given $\epsilon>0$ there exists $\delta_1>0$ such that;

$$|t-t_0|< \delta_1 \Rightarrow \|\vec{r}(t) - \vec{L}\|<\epsilon$$

You want to show that given $\epsilon$ there exists $\delta$ such that;

$$|t - t_0|< \delta \Rightarrow \left| \|\vec{r}(t)\|-\|\vec{L}\| \right|$$

Use reverse-triangle inequality i.e $\left| \|\vec{r}(t)\| - \|\vec{L}\| \right| \leq \|\vec{r}(t)- \vec{L}\|$. Therefore, choosing $\delta = \delta_1$ then we have;

$$|t-t_0| < \delta \Rightarrow \left| \|\vec{r}(t)\| - \|\vec{L}\| \right| \leq \|\vec{r}(t)-\vec{L}\| < \epsilon \Rightarrow \lim\limits_{t \to t_0} \|\vec{r}(t)\| = \|\vec{L}\|$$