Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $\alpha \in R$. Prove that a function $f:R\to R$ is differentiable at $\alpha$ if and only if there is a function $g:R\to R$ which is continuous at $\alpha$ and satisfies $f(x)-f(\alpha )=g(x)(x-\alpha )$ for all $x\in R$.

Carathéodory's Criterion

Welcome, fellow traveler on the path of mathematical discovery. Today, we are going to peel back the curtain on one of the most elegant, yet often overlooked, gems in calculus: Carathéodory's Criterion.

If you have ever felt that the standard definition of a derivative—that intimidating limit of a difference quotient—felt a bit clunky, you are not alone. Today, we will see how we can replace that limit with the smooth, intuitive language of continuity.

Imagine you are standing on a curve defined by a function $f(x)$. You pick a fixed point $\alpha$ and an arbitrary point $x$.

The slope of the secant line connecting these two points is given by the difference quotient:

Now, what if we defined a new function, $g(x)$, to be exactly this slope? That is, for $x eq \alpha$, we set $g(x)=\frac{f(x)-f(\alpha )}{x-\alpha }$.

If we rearrange this, we get the beautiful identity:

This is the heart of Carathéodory's Criterion. It tells us that $f$ is differentiable at $\alpha$ if and only if we can find a function $g$ that is continuous at $\alpha$ and satisfies this equation. It turns a problem of 'limits' into a problem of 'continuity.'

From Differentiability to Continuity

Let us assume $f$ is differentiable at $\alpha$. By definition, the derivative $f^{\prime }(\alpha )$ exists.

We construct our function $g(x)$ as the slope of the secant line for $x eq \alpha$. But what about at $x=\alpha$? To make $g$ continuous, we must define $g(\alpha )$ such that ${\lim }_{x\to \alpha }g(x)=g(\alpha )$.

Since

we simply define $g(\alpha )=f^{\prime }(\alpha )$. By this construction, $g$ is continuous at $\alpha$. We have successfully built a bridge from differentiability to continuity!

The Logical Reversal

Now, let us walk the bridge in the other direction. Suppose we are given a function $g$ that is continuous at $\alpha$, satisfying $f(x)-f(\alpha )=g(x)(x-\alpha )$.

We want to prove $f$ is differentiable at $\alpha$. We look at the difference quotient again:

As we take the limit as $x\to \alpha$, the left side becomes the definition of the derivative $f^{\prime }(\alpha )$. On the right side, because we assumed $g$ is continuous, the limit of $g(x)$ is simply $g(\alpha )$.

Therefore, the limit exists, and $f^{\prime }(\alpha )=g(\alpha )$. The proof is complete!

I know this might seem like a lot of work just to redefine the derivative. But think about the implications.

In many proofs—like the Chain Rule—we often struggle with the fact that the difference quotient is undefined at the point of interest. Carathéodory's Criterion allows us to bypass that singularity by working with the continuous function $g(x)$.

It is a reminder that in mathematics, the right perspective can turn a mountain of complexity into a gentle, rolling hill. Keep exploring, keep questioning, and remember: every complex limit is just a hidden, beautiful, continuous function waiting to be found.