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I have the following statement in my lecture slides:

For $f:E\rightarrow \mathbb{R}$, $E \subseteq \mathbb{R}$ and $c\in E $, if $\exists(x_n)$ in $E$ such that $x_n \ne c \ \forall n \in \mathbb{N}$ and $lim_{n\rightarrow \infty}(x_n)=c$ then $f'(c)$ can be defined.

I've attached the snippet from the slide: enter image description here

This seems incorrect given that the statement has used nothing about $f$ to reach a conclusion about the existence of its derivative at $c$. Furthermore, for any $c \in \mathbb{R}$ we can we define $(x_n)=c+1/n$ to satisfy these conditions. If there statement is incorrect is there some context or amendment under which it is correct?

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  • $\begingroup$ For $f'(c)$ to be well defined, we require every $(x_n)$ as described satisfy the condition:"$[f(x_n)-f(c)]/(x_n-c)$ converge to the same limit $L$", this $L$ is actually $f'(c)$ if the above is satisfied. $\endgroup$ – Li Chun Min May 2 '17 at 13:55
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I think this is an issue to do with terminology. I think that the statement is saying that if such a sequence exists, then we can consider the derivative of $f$ at $c$. It is true, as you noted, that the existence of such a sequence does not imply that the derivative of $f$ at $c$ is a well defined real number. But without such a sequence, we can't even begin to discuss whether or not the derivative exists.

It is not true that such a sequence exists for every $E \subset \mathbb{R}$ and $c \in E$. For example, $E=[0,1] \cup \{2\}$. Any sequence $(x_n)$ in $E$ such that $x_n \rightarrow 2$ must eventually be equal to $2$. The existence of such a sequence garuntees that $c$ is not isolated.

I do agree with you that the language used is confusing and potentially mislead. If you have the opportunity, I recommend talking to the writer of the notes and asking for the meaning to be clarified.

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  • $\begingroup$ Had a chance to contact my lecturer and indeed the interpretation is "if c can be approached by a sequence then it makes sense to investigate the derivative of a function at c." $\endgroup$ – Hamza May 2 '17 at 17:33
  • $\begingroup$ I am glad that they clarified $\endgroup$ – michaelhowes May 2 '17 at 21:06

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