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 \(\displaystyle \lim_{x \rightarrow a} f(x) = b \) then for every sequence \(\displaystyle (x_k)_{ k \in \mathbb{N} }\) with \(\displaystyle \lim_{k \rightarrow \infty } x_k = a\) we have \(\displaystyle \lim_{k \rightarrow \infty } f( x_k ) = b\) ... ...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.

 

[math771]

Hi Peter,

I can understand your confusion with this statement. "Obvious" is a subjective term and what may seem obvious to the authors may not be obvious to all readers. In this case, they are referring to the fact that for a function to be continuous at a point a, the limit of the function must exist at that point and be equal to the value of the function at that point.

To prove this, we can use the epsilon-delta definition of continuity. Let \epsilon > 0 be given. Since \lim_{x \rightarrow a} f(x) = b, we know that there exists a \delta > 0 such that for all x \in \mathbb{R}, if 0 < |x - a| < \delta, then |f(x) - b| < \epsilon.

Now, let (x_k)_{k \in \mathbb{N}} be a sequence with \lim_{k \rightarrow \infty} x_k = a. This means that for any \delta > 0, there exists a natural number N such that for all k > N, we have |x_k - a| < \delta. Therefore, for all k > N, we have |f(x_k) - b| < \epsilon, which implies that \lim_{k \rightarrow \infty} f(x_k) = b.

Thus, we have shown that for any 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. This completes the proof of Lemma 1.3.3.

I hope this helps clarify the "obvious" statement in the proof. Let me know if you have any other doubts.