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Let $\lim a_n=\lim b_n=c$ and $a_n\neq b_n$, and $f\in C(\Bbb R)$, then $\lim\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(c)$?

This question arise of an exercise where $a_n<b_n$ for all $n\in\Bbb N$ and $f\in C(\Bbb R)$. Then, by the mean value theorem we can write that

$$\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(\xi_n),\quad \xi_n\in(a_n,b_n)\tag{1}$$

Then, if Im not wrong, I can write

$$\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=\lim_{n\to\infty}f'(\xi_n)=f'(c)\tag{2}$$

because by the squeeze theorem $\lim a_n=c\le \lim \xi_n\le \lim b_n=c$ then $\lim \xi_n=c$.

Now suppose that we relax the initial conditions to $a_n\neq b_n$ then we can says this time that

$$\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(c)\tag{3}$$??

I can re-write (1) this time as

$$\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(\xi_n),\quad \xi_n\in(\min\{a_n,b_n\},\max\{a_n,b_n\})\tag{4}$$

but I cant apply again the squeeze theorem so I dont know how to justify (if it is justifiable) that (3) holds. Then my question: the statement on (3) holds? If so, how to justify it?

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  • $\begingroup$ If the derivative is not continuous at $c$, you can't deduce (2). $\endgroup$ – SSepehr Jan 24 '17 at 17:20
  • $\begingroup$ @SSepehr good point, but I can deduce something? By example that $\lim f'(\xi_n)$ is finite or so? I will update the question, thank you. $\endgroup$ – Masacroso Jan 24 '17 at 17:22
  • $\begingroup$ Is your function differentiable or just continuous? If it is just continuous, the statement is NOT true. $\endgroup$ – Mercy King Jan 24 '17 at 17:56
  • $\begingroup$ @MercyKing It is continuously differentiable in $\Bbb R$. $\endgroup$ – Masacroso Jan 24 '17 at 18:13
  • $\begingroup$ @Masacroso Why do you have $f\in C(\mathbb{R})$ in your statement, instead of $f \in C^1(\mathbb{R})$? $\endgroup$ – Mercy King Jan 24 '17 at 19:17
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Assuming $f'$ is continuous at $c$, then your argument works, and from the mean value theorem we have $$\lim_{n \rightarrow \infty} \frac{f(b_n) - f(a_n)}{b_n-a_n} =\lim_{n\to\infty}f'(\xi_n)=f'(c).$$ Also, $a'_n=\min\{a_n,b_n\}$ and $b'_n=\max\{a_n,b_n\}$ both converge to $c$, since for a given $\epsilon >0$ ,There's $N \in \mathbb{N}$ such that for all $n \ge N$ we have $|a_n - c| < \epsilon$ and $|b_n - c| < \epsilon$.

Note that if $f'$ is not continuous at $c$, the limit may not converge, and if it converges, it won't necessarily converge to $f'(c)$. For example, consider $f(x) = x^2 \sin(\frac{1}{x})$ where $f(0)$ is defined to be $0$. Then for $c = 0$, $\large a_n = \frac{1}{2\pi n+\frac{\pi}{2}}$, $\large b_n = \frac{1}{2\pi n}$, $$\lim_{n \rightarrow \infty} \frac{f(b_n) - f(a_n)}{b_n-a_n} =-\frac{2}{\pi} \ne f'(0)=0.$$

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    $\begingroup$ On the other hand, if $a_n \le c \le b_n$ and $a_n < b_n$ for all $n$ then the limit is $f'(c)$ even if $f$ is differentiable only at $c$ and at no other points. $\endgroup$ – Umberto P. Jan 24 '17 at 17:57

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