Limits of functions .... D&K Lemma 1.3.3 .... another question ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of Lemma 1.3.3 ...

Duistermaat and Kolk"s proof of Lemma 1.3.3. reads as follows:View attachment 7681In the above proof we read:

" ... ... The necessity is obvious ... ... "

Presumably this means that if $$\lim_{x \rightarrow a} f(x) = b $$ then for every sequence $$(x_k)_{ k \in \mathbb{N} }$$ with $$\lim_{k \rightarrow \infty } x_k = a$$ we have $$\lim_{k \rightarrow \infty } f( x_k ) = b$$ ... ...Although D&K reckon that it is obvious I cannot see how to (rigorously) prove the above statement ...

Can someone please demonstrate a rigorous proof ...
Help will be much appreciated ... ...

Peter

 

[castor28]

Hi Peter,

Your interpretation is correct (I just answered that in you previous thread before I read this one).

The point is that limits are not defined in exactly the same way for functions and sequences.

$\lim_{x\to a}f(x)=b$ means that for any $\varepsilon>0$ we can find $\delta>0$ such that $|x-a|<\delta\Rightarrow|f(x)-b|<\varepsilon$.

$\lim_{k\to\infty}x_k=a$ means that, for every $\delta>0$, there exists a $k_0\in\mathbb{N}$ such that $k>k_0\Rightarrow|x_k-a|<\delta$.

Now, for $k>k_0$, we have $|x_k-a|<\delta$, and, because of the hypothesis, $|f(x_k)-b|<\varepsilon$; this is precisely what $\lim_{k\to\infty} f(x_k) = b$ means.