I Definition of Limit for vector fields

[Kashmir]

Apostol defines limit for vector fields as
> $\quad \lim _{x \rightarrow a} f(x)=b \quad(\rm or\; f(x) \rightarrow b$ as $x \rightarrow a)$
means that :
$\lim _{\|x-a\| \rightarrow 0}\|f(x)-b\|=0$

Can't we say it's equivalent to $\lim _{x \rightarrow a}(f(x)-b)=0$

 

[BvU]

Scalars? vectors ?

$\$

 

[willem2]

Apostol defines limit for vector fields as
> $\quad \lim _{x \rightarrow a} f(x)=b \quad(\rm or\; f(x) \rightarrow b$ as $x \rightarrow a)$
means that :
$\lim _{\|x-a\| \rightarrow 0}\|f(x)-b\|=0$

Can't we say it's equivalent to $\lim _{x \rightarrow a}(f(x)-b)=0$

You can, after you accept the definition. I think going from f(x)= b to f(x)-b is 0, is obvious, but going from lim(x->a) to lim (||x-a|| ->0) is not.
There are certainly functions where it's important that || f(x) - b || < epsilon for all the x with || x-a || < delta.

 

[fresh_42]

Apostol defines limit for vector fields as
> $\quad \lim _{x \rightarrow a} f(x)=b \quad(\rm or\; f(x) \rightarrow b$ as $x \rightarrow a)$
means that :
$\lim _{\|x-a\| \rightarrow 0}\|f(x)-b\|=0$

Can't we say it's equivalent to $\lim _{x \rightarrow a}(f(x)-b)=0$

Say $\lim_{x \to a} f(x)=b$ means, that for each neighborhood $V_b$ of $b$ there is a neighborhood $U_a$ of $a$ such that $f(x)\in V_b$ whenever $x\in U_a\,.$ If we further assume that we are dealing with topological vector spaces, then this is equivalent to $\lim_{x-a \to 0}(f(x)-b)=0\,.$ It also implies that every neighborhood contains an $\varepsilon -$ball. Thus we can replace all neighborhoods with $\varepsilon -$balls, i.e. the norm convergence. And metric spaces are topological vector spaces.

 

[Kashmir]

Scalars? vectors ?

$\$

I couldn't understand. Could you please explain what you mean?

 

[Kashmir]

You can, after you accept the definition. I think going from f(x)= b to f(x)-b is 0, is obvious, but going from lim(x->a) to lim (||x-a|| ->0) is not.
There are certainly functions where it's important that || f(x) - b || < epsilon for all the x with || x-a || < delta.

"lim(x->a) to lim (||x-a|| ->0)" because if x->a the magnitude of difference vector ||x-a|| will tend to zero.
Is that ok?

 

[Kashmir]

Say $\lim_{x \to a} f(x)=b$ means, that for each neighborhood $V_b$ of $b$ there is a neighborhood $U_a$ of $a$ such that $f(x)\in V_b$ whenever $x\in U_a\,.$ If we further assume that we are dealing with topological vector spaces, then this is equivalent to $\lim_{x-a \to 0}(f(x)-b)=0\,.$ It also implies that every neighborhood contains an $\varepsilon -$ball. Thus we can replace all neighborhoods with $\varepsilon -$balls, i.e. the norm convergence. And metric spaces are topological vector spaces.

Thank you for your reply. I can't understand it right now. It's beyond my learning. Hopefully I'll come back to it after studying some more. :)