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I have read that the derivative of a function $f$ at a point $a$ is given by $f'(a) = \lim_{h \to 0}((f(a+h)-f(a))/h)$ provided that $f$ is continuous at $a$ (or in other words, that $\lim_{x \to a}(f(x))=f(a)$).

One gripe that I have with this definition is that the possible output values of the limit are $\mathbb{R} \cup \{\infty, -\infty\}$, meaning that the possible values of the derivative are also $\mathbb{R} \cup \{\infty, -\infty\}$. I don't think it makes much sense for the derivative to equal infinity at a point where the tangent line is vertical for a single point on the curve. The derivative is supposed to measure the slope of the tangent line, and a vertical tangent line has no defined slope. One such example is the function $f(x) = \sqrt{x}$, which is continuous at $0$, and whose derivative at $0$ is equal to $\infty$. Both of those can be proven by using delta-epsilon techniques.

What is needed is a "cast" from the output value of the limit to the real numbers before assigning it to the derivative. For example, we should say that if $\lim_{h \to 0}((f(a+h)-f(a))/h) \in \mathbb{R}$, then $f'(a) = \lim_{h \to 0}((f(a+h)-f(a))/h)$ provided that $\lim_{x \to a}(f(x))=f(a)$.

J. W. Tanner
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Lauren S
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    What is the derivative of $\sqrt x$ at $x=0?$ – Thomas Andrews Apr 21 '24 at 17:55
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    The fact that a function that is differentiable at $a$ must be continuous at $a$ is a consequence of differentiability, not a requirement. And when a limit "equals" $\pm\infty$, the limit does not exist (we are just saying why it does not exist). – Arturo Magidin Apr 21 '24 at 18:01
  • Some quick observations: (1) Continuity does not need to be assumed in the definition of a derivative. (2) Whether (signed) infinite values can be the result of a limit varies with author and context. (3) The notion of infinite derivatives certainly arises in mathematics -- see The set of points where a function has infinite derivative and the references that links in the comments lead you to. Of course, they don't start appearing until at least upper undergraduate and beginning graduate level analysis (well past calculus). – Dave L. Renfro Apr 21 '24 at 18:03

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The last paragraph is how it is actually defined. You say $f$ is differentiable at $a$ if that limit exists, and only then define the derivative $f^\prime(a)$. Only continuity of $f$ does not ensure that the limit exists (even if you include $\infty$), for example $f(x)=|x|$.

Hope this helps. :)

ultralegend5385
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