# Thread: Another Derivative Problem

1. ## Another Derivative Problem

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

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

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

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 $\frac{f(x_n)-f(z_n)}{x_n-z_n} = \frac{f(x_n)-f(a)}{x_n-a} \frac{x_n-a}{x_n-z_n} + \frac{f(z_n)-f(a)}{z_n-a} \frac{a-z_n}{x_n-z_n}$, show that this lie between $\frac{f(x_n)-f(a)}{x_n-a}$ and $\frac{f(z_n)-f(a)}{z_n-a}$.

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

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

5. May be you can use this

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

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

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

I hope that this is correct. Any opinion?

7. Or you could say that

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

and then

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

Or vice versa.