0

I have just shown that if a function $f:(a,b)\rightarrow \mathbb{R}$ is differentiable at $c\in (a,b)$, and $(x_n),(y_n)\subset(a,b)$ are two sequences that tend to c, with $x_n\neq y_n$ and $(*)$ $x_n\leq c\leq y_n$ $\forall n$, then $$\lim \limits_{x \to c} \frac{f(y_n)-f(x_n)}{y_n-x_n}\; \text{ exists.}$$

My proof uses the fact that $f$ differentiable at c $\iff$ $\exists$ L, $\epsilon(x)$ on $(a,b)$ such that $$f(x)=f(c)+L(x-c)+\epsilon(x)(x-c)$$ where $\epsilon \to 0$ as $x\to c$. Namely $L=f'(c)$ and $\epsilon(x)=\frac{f(x)-f(c)}{x-c}-L$

Using this we can write $\frac{f(y_n)-f(x_n)}{y_n-x_n}=L+\frac{\epsilon(y_n)(y_n-c)-\epsilon(x_n)(x_n-c)}{y_n-x_n}$. The modulus of the second expression on the right hand side is less than or equal to

$$\lvert \frac{\epsilon(y_n)(y_n-c)}{y_n-x_n}\rvert+\lvert \frac{\epsilon(x_n)(x_n-c)}{y_n-x_n}\rvert \leq \lvert\frac{\epsilon(y_n)(y_n-x_n)}{y_n-x_n}\rvert+\lvert\frac{\epsilon(x_n)(y_n-x_n)}{y_n-x_n}\rvert$$ firstly using the triangle inequality and then using $(*)$. Then the fact $\epsilon \to 0$ as $x\to c$ completes this (the limit is $f'(c)$).

I am wondering if this result holds still if $(*)$ is no longer assumed? I haven't been able to prove that it does but I still haven't found a valid counterexample?

jcneek
  • 714
  • 2
  • 13
  • Check this: https://math.stackexchange.com/q/2796804/42969 – Martin R May 15 '21 at 18:05
  • @MartinR Thank you for this. I think the given answer is focusing more on the limit existing but not equalling zero? Also I don't understand the equality here " ${{f(x_n)-f(y_n)}\over{x_n-y_n}}=f'(c_n)(x_n-y_n)$ ". Could you clarify this? – jcneek May 15 '21 at 20:14
  • @jaycneek That is the mean value theorem – Mason May 15 '21 at 22:03
  • @Mason wouldn't that be $$\frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(c_n)$$ – jcneek May 15 '21 at 22:16
  • @jaycneek Yes. $(c_n) \to c$, so if $f'$ is not continuous at $c$, the equality will not hold. – Mason May 16 '21 at 00:01

0 Answers0