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Exercise: Let $f : [a, b] \to \mathbb{R}$ be continuous on the closed, bounded interval $[a,b]$. Also assume that $f $ is differentiable on the open interval $(a, b)$. Assume that there exists a number $ c \in \mathbb{R}$ such that $f’(x) \to c$ for $x \to$ $a_+$ . Show that $f$ is differentiable at the endpoint $a$ with $f’(a) = c$.

My solution (which is not correct according to professor):

In the definition of continuity in the point $a$ $\Delta x \in (0, b-a)$. We are asked to show: \begin{align} &\frac{\Delta f}{\Delta x} = \frac{\Delta f}{x-a}\rightarrow c, \hspace{3mm} \Delta x= x-a \rightarrow 0 \Leftrightarrow f'(a) = c\\ &\Rightarrow \frac{\Delta f}{x-a} \rightarrow c, \hspace{3mm} x \rightarrow a_+ \label{eq: differentiabilitet_specielt} \end{align} But since the first equation holds for $ x \in [a,b] \setminus \{ a\}$, the left implication is specially true in point a. Which means the two expressions are equvialent for our case.

I would like to understand why my solution is incorrect.

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1 Answers1

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Let $x\in(a,b)$ and by the mean value theorem there exists some $y\in(a,x)$, s.t. $$\frac{f(x)-f(a)}{x-a}=f'(y)$$Now let $x\to a$, which implies $y\to a$ and we get $$\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim\limits_{y\to a}f'(y)=c$$i.e. f is differentiable in a and $f'(a)=c$.

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