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Let $f: [a,b] \to \mathbb{R}$ be a map that is differentiable at a point $x \in [a,b]$. Then $f$ is continuous at $x$.

I'm trying to be very rigorously here. For reference, here's the definition of differentiability I'm using:

Let $f$ be defined (and real-valued) on $[a,b]$. For any $x \in [a,b]$, form the quotient $$\phi(t) = \frac{f(t) - f(x)}{t-x}\quad (a < t <b, t \neq x)$$ and define $$f'(x) = \lim_{t \to x} \phi(t)$$

Proof:

It suffices to show that $\lim_{t \to x, t \in [a,b]} f(t) = f(x)$

And, $$\lim_{t \to x, t \in [a,b]} f(t) = \lim_{t\to x, t \in [a,b]} [(f(t)-f(x))+f(x)]$$ $$= \lim_{t \to x, t \in (a,b), t \neq x} [(f(t)-f(x))+f(x)]$$ $$= \lim_{t \to x, t \in (a,b), t \neq x} \left[\frac{f(t)-f(x)}{t-x}(t-x) + f(x)\right]$$ $$= \underbrace{\lim_{t \to x, t \in (a,b), t \neq x} \frac{f(t)-f(x)}{t-x}}_{\in \mathbb{R}, by \ assumption}\underbrace{\lim_{t \to x, t \in (a,b), t \neq x}(t-x)}_{= 0} + f(x) = f(x)$$

Is everything I did correct? Nobody ever writes down these limit subranges. Why is that?

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    Looks correct. There was a similar post yesterday: https://math.stackexchange.com/a/2865578/168433 – md2perpe Jul 29 '18 at 19:53
  • In the case at hand the subranges are "tacitly understood". During a limit $t\to x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway. – Christian Blatter Jul 30 '18 at 07:49

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