Differentiability implying continuity

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 $x=1$?

Solution 1:

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 $\lim\limits_{x\to c}\,f(x)=f(c)$.

Solution 2:

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.