let $f: J \to \mathbb{R}$ . let $x_n < c <y_n$ be such that $(y_n-x_n) \to0$ show that $\lim _ {n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(c)$

Question

let $f: J \to \mathbb{R}$ be differential function . let $x_n < c <y_n$ be such that $(y_n-x_n) \to0$

show that

$$\lim _ {n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(c)$$

i am trying to apply Mean value theorem on $f$ on the interval $[y_n,x_n]$.this gives

$$\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(\theta_n)$$

Now as $n$ goes to infinity $\theta_n \to c$. but $f'$ is not given to be continuous.if that was given then we had done.

But how to approach this problem since $f'$ is not given to be continous . Any Hint.

Answer 1

$\frac {f(y_n)-f(x_n)} {y_n-x_n}-f'(c)=\frac {y_n -c} {y_n-x_n} [\frac {f(y_n)-f(c)} {y_n-c}-f'(c)]+\frac {c-x_n} {y_n-x_n} [\frac {f(c)-f(x_n)} {c-x_n}-f'(c)]$. Can you complete the proof now?