Let $f: \Bbb[a,b]\rightarrow\Bbb{R}$ be differentiable in a point $a<x_0<b$. Use Taylor's expansion with the remainder of $f$ around $x_0$ to show that if $ x_n\rightarrow\ x_0^{-}$, and $ y_n\rightarrow\ x_0^{+}$ then: $$ \lim \limits_{n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0)$$

I know that:

$f(x_n)= f(x_0)+f^\prime(x_0)(x_n-x_0) + R_1 (x_n) $


$f(y_n)= f(x_0)+f^\prime(x_0)(y_n-x_0) + R_1 (y_n)$

Note: $R_1$ is the remainder.

and then by substracting:

$ \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0) + \frac{R_1(x_n)-R_1(y_n)}{x_n-y_n}$

but It seemed that I couldn't continue from here to anywhere, can someone please show the right way to do this?

  • 2
    $\begingroup$ Hint: Using the fact that $x_n\lt x_0\lt y_n$ for $n$ large enough, one gets $$\left|\frac{R_1(x_n)-R_1(y_n)}{x_n-y_n}\right|\leqslant\left|\frac{R_1(x_n)}{x_n-x_0}\right|+\left|\frac{R_1(y_n)}{y_n-x_0}\right|.$$ $\endgroup$
    – Did
    Mar 5 '15 at 13:56
  • $\begingroup$ @Did and this means that both expressions on the right side converge to zero ? :) $\endgroup$
    – Xhero39
    Mar 5 '15 at 14:00

A nice exercise in Rudin's Principles of Mathematical Analysis (Exercise 5.19) states that, if $\alpha_n \to 0$ and $\beta_n \to 0$, then $$ D_n = \frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n} $$ converges to $f'(0)$ under each of the following assumptions:

  1. $\alpha_n < 0 <\beta_n$;
  2. $0<\alpha_n<\beta_n$ and $\{\beta_n/(\beta_n-\alpha_n)\}_n$ is bounded;
  3. $f'$ is continuous.

Here you are in case 1. You can write $$ D_n = \frac{f(\beta_n)-f(0)}{\beta_n-0}\frac{\beta_n}{\beta_n-\alpha_n} + \frac{f(\alpha_n)-f(0)}{\alpha_n}\frac{-\alpha_n}{\beta_n-\alpha_n}. $$ Then you remark that $0<\frac{\beta_n}{\beta_n-\alpha_n}<1$ and $-1<\frac{\alpha_n}{\beta_n-\alpha_n}<0$. You should find your way from here.

By the way, case 3 is easily covered by some MVT or Taylor's theorem.


You need to use an estimate for the remainder. For exemple

$$R_1(x) = (x-x_0)\epsilon(x)$$

with $$\lim_{x\to x_0}\epsilon(x) = 0$$

Or in other words, $R_1(x) = o(x-x_0)$

  • $\begingroup$ How can you assume that $f''$ exists? $\endgroup$
    – Siminore
    Mar 5 '15 at 13:35
  • $\begingroup$ Indeed, not the right formula for the remainder. Will correct $\endgroup$
    – Tryss
    Mar 5 '15 at 13:47
  • 1
    $\begingroup$ Where did you use the fact that $x_n < y_n$? $\endgroup$
    – Siminore
    Mar 5 '15 at 13:54
  • $\begingroup$ ok, and I get this: $\frac{R_1(x_n)-R_1(y_n)}{x_n-y_n}$ = $\frac{o(x_n-x_0)-o(y_n-x_0)}{(x_n-x_0)+(y_n+x_0)}$ now I need to prove that this things tends to zero. How? $\endgroup$
    – Xhero39
    Mar 5 '15 at 13:56
  • $\begingroup$ better to use $R_1(x_n) = (x_n-x_0)\epsilon(x_n)$. $\endgroup$
    – Tryss
    Mar 5 '15 at 14:02

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