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Theorem: If $f$ is finitely differentiable at $c$, then $f$ is also continuous at $c$. But according to the definition, for a function to be differentiable at $c$, it need not even be defined at $c$, just that it should have a finite value in the vicinity of $c$.

My question: If function is not defined at $c$, how can it be continuous at $c$? For example, say we have a straight horizontal line function which is broken at $x=1$, the function is defined at all the points except $x=1$, it has a finite value at all the points except $x=1$, since it is having a value in the nearby region of $x=1$ we say that it is differentiable at $x=1$ but can we say that it is continuous at $x=1$?

Davide Giraudo
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Vikram
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2 Answers2

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It is not true that a function that is differentiable at $c$ need not be defined at $c$. Note that differentiability means existence of the limit $\displaystyle{f'(c)=\lim\limits_{h\to 0}\,\frac{f(c+h)-f(c)}{h}}$, or, equivalently, $\displaystyle{f'(c)=\lim\limits_{x\to c}\,\frac{f(x)-f(c)}{x-c}}$, and for this to make sense $f(c)$ must be defined.

If $g$ is a function defined near $c$, then it is possible for $\lim\limits_{x\to c}\,g(x)$ to exist, even if $g$ is not defined at $c$. If $g$ is the horizontal line function with a gap at $c$, this would occur. But the derivative is a limit of a quotient that depends on $f(c)$. The expression $g(x)=\displaystyle{\frac{f(x)-f(c)}{x-c}}$ itself is not defined when $x=c$, but if the limit exists, then $\displaystyle{\lim\limits_{x\to c}\,g(x)=\lim\limits_{x\to c}\,\frac{f(x)-f(c)}{x-c}=f'(c)}$.

A standard way to show that differentiability implies continuity is to note that for $x\neq c$, $f(x)-f(c)=(x-c)\frac{f(x)-f(c)}{x-c}$, take the limit of both sides, use the fact that the limit of a product is the product of the limits, and conclude that $\lim\limits_{x\to c}\,f(x)=f(c)$.

Jonas Meyer
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  • Mr. Meyer, how would we adjust this proof to cover the case of complex functions defined on open sets? (Bonus: why open sets rather than closed?) – The Chaz 2.0 Feb 14 '12 at 22:23
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    The Chaz, I do not understand the question. If $f:A\to\mathbb C$ where $A\subseteq \mathbb C$ and $c$ is both an element of $A$ and a limit point of $A$ (whether or not $A$ is open), then it makes sense to ask whether $\lim\limits_{z\to c,z\in A}\frac{f(z)-f(c)}{z-c}$ exists, and if it does we can call it $f'(c)$. If $f'(c)$ exists, then $f$ is continuous at $c$ for the same reason as in the real case. – Jonas Meyer Feb 14 '12 at 23:12
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When we say a function is continuous (or differentiable), what we mean is that at each point for which the function is defined the function is continuous (or differentiable). So the points at which a function is not defined do not come into play for either. The example you gave is a indeed a continuous function on $\mathbb{R}\setminus \{0\}$, which is its domain, that is the set of points on which it is defined.

Alex Becker
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