Exercise 788Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL2Added 2 Nov 2026

The limit of the derivative theorem

#mean value theorem#derivative at a point#extension

Statement

Let $f$ be continuous on $[a,b)$ and differentiable on $(a,b)$ , and suppose $f'(x)\to\ell\in\mathbb{R}$ as $x\to a^+$ . Prove that $f$ is differentiable at $a$ from the right, with $f'_+(a)=\ell$ .

Then apply this to $f(x)=e^{-1/x^2}$ for $x\ne0$ , $f(0)=0$ , to show that $f$ is differentiable at $0$ with $f'(0)=0$ .

Hint

The difference quotient $\dfrac{f(x)-f(a)}{x-a}$ is exactly what the mean value theorem handles.

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Solution

Proof. For $x\in(a,b)$ , the mean value theorem on $[a,x]$ gives $c_x\in(a,x)$ with
$\frac{f(x)-f(a)}{x-a}=f'(c_x).$
As $x\to a^+$ , $c_x\to a^+$ so $f'(c_x)\to\ell$ . Hence the right derivative at $a$ exists and equals $\ell$ .

Application. $f$ is continuous at $0$ because $e^{-1/x^2}\to0$ . For $x\ne0$ ,
$f'(x)=\frac2{x^3}e^{-1/x^2}.$
With $u=\frac1{x^2}\to+\infty$ , $|f'(x)|=2|x|^{-3}e^{-u}=2u^{3/2}e^{-u}\to0$ (the exponential beats powers). So $f'(x)\to0$ at $0^\pm$ , and by the theorem (applied on each side) $f'(0)=0$ .

Remark. This function is very flat at $0$ : all its derivatives vanish there. It is a famous example of a smooth function that is not equal to its Taylor series.

The limit of the derivative theorem

Statement

Solution

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