# Show that if $x_n\rightarrow\ x_0^{-}$ and $y_n\rightarrow\ x_0^{+}$ then $\lim \limits_{n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0)$

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:

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

and:

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

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?

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

• How can you assume that $f''$ exists? Mar 5 '15 at 13:35
• Indeed, not the right formula for the remainder. Will correct Mar 5 '15 at 13:47
• Where did you use the fact that $x_n < y_n$? Mar 5 '15 at 13:54
• 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? Mar 5 '15 at 13:56
• better to use $R_1(x_n) = (x_n-x_0)\epsilon(x_n)$. Mar 5 '15 at 14:02