# Harder limits

• February 4th 2010, 12:30 PM
vuze88
Harder limits
The function f is differentiable at a. Find $\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}$

I'd assume its something to do with manipulating the definition $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ but have no idea how.
• February 4th 2010, 05:49 PM
felper
Quote:

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

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

$\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\$

could you finish it?
• February 4th 2010, 07:23 PM
vuze88
so is the answer $2pf'(a)$
• February 5th 2010, 06:15 AM
felper
Quote:

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

could you finish it?

$p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}$
$=p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a)-f(a-ph)}{ph}$
$=p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a-ph)-f(a)}{-ph}=2pf'(a)$