If $f$ is differentiable for $x\neq x_0$ and $\lim_{x\to x_0} f'(x) = c$ then $f'(x_0) = c$

Question

Exercise :

Let $f : \mathbb R \to \mathbb R$ and $x_0 \in \mathbb R$. Suppose that $f$ is differentiable for all $x \neq x_0$. If $\lim_{x \to x_0}f'(x) = c \in \mathbb R$ show that $f$ is differentiable at $x_0$ and $f'(x_0) = c$.

Attempt :

Isn't it pretty straight forward that since $\lim_{x \to x_0} f'(x) = c$ then $f'$ is continuous at $x_0$ and thus differentiablewith $f'(x_0) = c$ ? Does it need some more delicate or rigorous mathematical proof ?

There certainly is some function $g$ such that $g$ is continuous at $x_0$ and $g(x)=f'(x)$ for all $x\ne x_0$ and $g(x_0)=c$. Now I claim that $g(x_0)$ is not the derivative of $f$ at $x_0$; prove me wrong. — , Sep 06 '18 at 00:32
No one ever told you anything about continuity of $f'$ ... and it can't be continuous at $x_0$ if it's not defined there. ... Serious hint: Try using the Mean Value Theorem. — Ted Shifrin, Sep 06 '18 at 00:34
Possible duplicate of If $f'(x)$ has a limit as $x\to x_0$, then the function $f$ is differentiable at $x_0$ — Theo Bendit, Sep 06 '18 at 00:44
You missed to state that $f$ is given continuous at $x_0$. Without this assumption the result is wrong. — Paramanand Singh, Sep 06 '18 at 02:25
It is not straightforward that $\lim_{x\to x_0}f'(x)=c$ implies $c=f'(x_0).$ — DanielWainfleet, Sep 06 '18 at 06:32
The exercise as stated is incorrect. We must also assume that $f$ is continuous at $x_0$. — DanielWainfleet, Sep 06 '18 at 06:38
@DanielWainfleet Since $f : \mathbb R \to \mathbb R$ and $x_0 \in \mathbb R$ then $f$ is continuous at $x_0$. — Rebellos, Sep 06 '18 at 06:51
@Rebellos no it's not. Take $$f(x) = \begin{cases}1, & x=0 \ 0, &x\not=0\end{cases}$$ and $x_0 = 0$ then $f: \Bbb R \to \Bbb R$ and $x_0 \in \Bbb R$ but still: $f$ is not continuous at $x_0$. Additionally this $f$ contradicts your question so you do need continuity at $x_0$. — Gono, Sep 06 '18 at 07:08
If $f$ is not differentiable at $x_0$ then you cannot use the notation $f'(x_0)$ it doesn't even exist. So you can't write that $f'(x_0)=c$, we must keep it in the form of the limit — Fareed Abi Farraj, Sep 06 '18 at 07:45
Possible duplicate of Prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$. — Hans Lundmark, Sep 06 '18 at 08:39
$f:\Bbb R\to \Bbb R$ means that $f$ is a function with domain and range $\Bbb R$. It does not mean that $f$ is continuous at $x_0$ or at any other point of $\Bbb R$ — DanielWainfleet, Sep 06 '18 at 09:53

score 1 · Accepted Answer · answered Sep 07 '18 at 15:17

1

Figured out that applying De L'Hospital provides a brutally fast proof (assuming $f$ is continuous at $x_0$) :

$$\lim_{x \to x_0}\frac{f(x) - f(x_0)}{x-x_0} = \lim_{x \to x_0} \frac{[f(x) - f(x_0)]'}{(x-x_0)'}=\lim_{x\to x_0} \frac{f'(x)}{1}=c \implies f'(x_0) = c$$

answered Sep 07 '18 at 15:17

Rebellos

21,324

1

That's a very important application of the theorem! But, as you see, it can be applied if $f$ is continuous at $x_0$. With an easy counterexample, we see that continuity is necessary anyhow. – egreg Sep 07 '18 at 15:25

score 0 · Answer 2 · answered Sep 06 '18 at 07:28

Assuming $f$ is continuous at $x_0.$ It is equivalent to show $g'(0)=c$ where $g(x)=f(x+x_0)-f(x_0).$ That is, to show $\lim_{x\to 0}\frac {g(x)}{x}=c.$

We have $\lim_{x\to 0}g'(x)=c$ and $\lim_{y\to 0}g(y)=g(0)=0.$

Since $\lim_{x\to 0}g'(x)=c,$ there exists $x_1>0$ such that $g'$ is bounded on $(0,x_1).$ For $x\in (0,x_1)$ let $M(x)=\sup_{z\in (0,x)}g'(z)$ and $m(x)=\inf_{z\in (0,x)}g'(z).$

For $0<y<x<x_1$ we have, by the MVT, $g(x)=g(y)+(x-y)g'(z_{x,y})$ for some $z_{x,y}\in (y,x).$ So $$\frac {g(x)}{x}=\frac {g(y)}{x}+(1-\frac {y}{x})g'(z_{x,y}).$$ Therefore $$\frac {g(y)}{x}+(1-\frac {y}{x})M(x)\geq \frac {g(x)}{x}\geq \frac {g(y)}{x}+(1-\frac {y}{x})m(x).$$

Now keeping $x$ fixed and letting $y\to 0,$ since $\lim_{y\to 0}g(y)=0,$ we have $$M(x)\geq \frac {g(x)}{x}\geq m(x)$$ for all $x\in (0,x_1)$.

Since $c=\lim_{x\to 0+}M(x)=\lim_{x\to 0+}m(x),$ we have therefore $ \lim_{x\to 0+}\frac {g(x)}{x}=c.$

Similarly, $\lim_{x\to 0-}\frac {g(x)}{x}=c.$

Remark . The change from $f$ to $g$ is not really necessary. It reduces what I call the "clutter", to make it easier to follow.

This is essentially the proof of L'Hôpital's theorem in a special case. — egreg, Sep 07 '18 at 15:27

If $f$ is differentiable for $x\neq x_0$ and $\lim_{x\to x_0} f'(x) = c$ then $f'(x_0) = c$

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