Could anyone help me with this proof?
Prove that $\lim_{x\to a} f(x)=0$ whenever $\lim_{x\to a} \|f(x)\|=0$ for $x\in \mathbb{R}^n$
Where $f(x)$ is a vector function and $\|f(x)\|$ is the vector norm.
My try:
I let $\|f(x)-0\|< \delta_1$ whenever $\|x-a\|< \epsilon_1$ for the first function.
$\|f(x)-0\|< \delta_2$ whenever $\|x-a\|< \epsilon_2$ for the second function.
Hence $f(x)< \|f(x)\| < \delta_2 =\epsilon_2$.
Is my proof correct? Could anyone explain? Thanks.
We are required to prove this result only when $ \lim_{x \to a}||f(x)|| = 0 $ and so we can assume this.
So Let $\epsilon \gt 0$ be an arbitrary quantity. We must now find a corresponding $\delta$ such that $ ||f(x) - 0 || \lt \epsilon$ whenever $||x - a|| \lt \delta$. Note that since a norm is non-negative,
$$ ||f(x) - 0 || = ||f(x) || = \left|{(||f(x) ||)}\right| = \left|{(||f(x) ||) - 0}\right| $$
Since we know that $ \lim_{x \to a}||f(x)|| = 0 $ for any $\epsilon \gt 0\ $ there is a $\delta \gt 0$ such that $$ ||x - a|| \lt \delta \implies \left|{(||f(x) ||) - 0}\right| \lt \epsilon \implies ||f(x) - 0 || \lt \epsilon$$
$\mathscr Q.E.D.$