1. ## Harder limits

The function f is differentiable at a. Find $\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}$

I'd assume its something to do with manipulating the definition $\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ but have no idea how.

2. Originally Posted by vuze88
The function f is differentiable at a. Find $\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}$

I'd assume its something to do with manipulating the definition $\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ but have no idea how.
$\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\$

could you finish it?

3. so is the answer $\displaystyle 2pf'(a)$

4. Originally Posted by felper
$\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\$

could you finish it?
$\displaystyle p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}$
$\displaystyle =p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a)-f(a-ph)}{ph}$
$\displaystyle =p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a-ph)-f(a)}{-ph}=2pf'(a)$