1. ## Another Derivative Problem

Let $\displaystyle f$ be differentiable at $\displaystyle a$ and let $\displaystyle {x_n}$ and $\displaystyle {z_n}$ be two sequences converging to $\displaystyle a$ such that $\displaystyle x_n < a < z_n$ for $\displaystyle n \in \mathbb{N}$. Prove that
$\displaystyle \lim_{n \to \infty} \frac{f(x_n)-f(z_n)}{x_n-z_n} = f^{'}(a)$.

2. This looks like a result of the Mean Value Theorem...

3. $\displaystyle x_n$ and $\displaystyle z_n$ converges to $\displaystyle a$ and $\displaystyle x_n<a<z_n$. This gives $\displaystyle lim_{n \to \infty} x_n = a = lim_{n \to \infty} z_n$. In other words, $\displaystyle lim_{n \to \infty}(x_n-z_n)=0$. Again, $\displaystyle lim_{n \to \infty} x_n = lim_{n \to \infty} z_n$; consequently, $\displaystyle lim_{n \to \infty} f(x_n) - lim_{n \to \infty} f(z_n)$ or $\displaystyle lim_{n \to \infty} (f(x_n) - f(z_n) )$ converges to $\displaystyle 0$.
Thus the expression can be thought of the form $\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ that is $\displaystyle f'(a).$

4. Hey,

The question doesn't allow us to use mean value theorem directly to the function.

Anyway,the hint given to this question is that using $\displaystyle \frac{f(x_n)-f(z_n)}{x_n-z_n} = \frac{f(x_n)-f(a)}{x_n-a} \frac{x_n-a}{x_n-z_n} + \frac{f(z_n)-f(a)}{z_n-a} \frac{a-z_n}{x_n-z_n}$, show that this lie between $\displaystyle \frac{f(x_n)-f(a)}{x_n-a}$ and $\displaystyle \frac{f(z_n)-f(a)}{z_n-a}$.

I've tried using this hint, but I can't get anywhere between $\displaystyle \frac{f(x_n)-f(a)}{x_n-a}$ and $\displaystyle \frac{f(z_n)-f(a)}{z_n-a}$. And if I'm not wrong, we can't assume that the function is monotonic as there might be cases where $\displaystyle f(x_n) \leq f(z_n)$ for some $\displaystyle n \in \mathbb{N}$ and $\displaystyle f(x_n) \geq f(z_n)$ for the remaining integers in $\displaystyle \mathbb{N}$.

So any idea on how to continue from the hint above?

5. May be you can use this

$\displaystyle f'_{max}(a)=max( \; \frac{f(x_n)-f(a)}{x_n-a}\; ; \; \frac{f(z_n)-f(a)}{z_n-a} \; )$

$\displaystyle f'_{min}(a)=min( \; \frac{f(x_n)-f(a)}{x_n-a}\; ; \; \frac{f(z_n)-f(a)}{z_n-a} \; )$

6. $\displaystyle \frac{f(x_n)-f(z_n)}{x_n-z_n}$
$\displaystyle = \frac{f(x_n)-f(a)}{x_n-a} \frac{x_n-a}{x_n-z_n} + \frac{f(z_n)-f(a)}{z_n-a} \frac{a-z_n}{x_n-z_n}$
$\displaystyle \leq max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) \frac{x_n-a}{x_n-z_n} + max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) \frac{a-z_n}{x_n-z_n}$
$\displaystyle = max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a})$.
Similarly,
$\displaystyle \frac{f(x_n)-f(z_n)}{x_n-z_n} \geq min(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a})$.
Then $\displaystyle min(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) \leq \frac{f(x_n)-f(z_n)}{x_n-z_n} \leq max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a})$.
Since $\displaystyle lim_{n \to \infty} x_n = a$ and $\displaystyle lim_{n \to \infty} z_n = a$,$\displaystyle lim_{n \to \infty} \frac{f(x_n)-f(a)}{x_n-a} = f'(a)$ and $\displaystyle lim_{n \to \infty} \frac{f(z_n)-f(a)}{z_n-a} = f'(a)$. Then $\displaystyle lim_{n \to \infty} max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) = f'(a)$ and $\displaystyle lim_{n \to \infty} max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) = f'(a)$. Then by squeeze theorem, $\displaystyle lim_{n \to \infty} \frac{f(x_n)-f(z_n)}{x_n-z_n} = f'(a)$.

I hope that this is correct. Any opinion?

7. Or you could say that

$\displaystyle f'_{max}(a)= \; \frac{f(z_n)-f(a)}{z_n-a}$

and then

$\displaystyle f'_{min}(a)= \; \frac{f(x_n)-f(a)}{x_n-a} .$

Or vice versa.