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A student of mine is in a class that uses the following definition of derivative: $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ Their definition of critical points is "places where the derivative of a function is either zero or undefined".

Yet they do not list the endpoints of a function on a closed interval to be critical points. IMO, based on the definition of derivative, the derivatives of the endpoints are undefined and therefore should count as critical points. Is there an inconsistency here, or am I missing something?

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    $\begingroup$ For the definition of $\lim_{h\to 0}$, they probably just require zero to be an accumulation point of the domain of the function (in $h$) to which the limit is applied. And, yes, the ends of a closed interval are accumulation points of the interval. $\endgroup$
    – user700480
    Oct 13, 2021 at 17:33
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    $\begingroup$ If the function is not defined on one side of $a$ and continuous on the other side of $a.$ e.g. $\lim{x\to 0} \sqrt x$ the limit still exists (and behaves like a right-hand limit in this example). $\endgroup$
    – Doug M
    Oct 13, 2021 at 18:59

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Although, in general, $\lim_{x\to a} f(x)$ only exists if $\lim_{x\to a^-} f(x) =\lim_{x\to a^+} f(x)$, and therefore $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ wouldn't exist at the endpoints, there are additional things to consider.

Usually, if the point at $h=0$ is an accumulation point in the domain of the function, then the limit is considered to exist. And the endpoints would meet this requirement.

In general practice, if a function is not defined on one side of a point and continuous on the other side, then the limit is still considered to exist, such as in $\lim_{x\to 0} \sqrt{x}$.

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