Harder limits

**vuze88** · February 4th 2010, 12:30 PM

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.

**felper** · February 4th 2010, 05:49 PM

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?

**vuze88** · February 4th 2010, 07:23 PM

so is the answer $2pf'(a)$

**felper** · February 5th 2010, 06:15 AM

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?

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

Thread: Harder limits

LinkBack

Thread Tools

Display

Harder limits

Similar Math Help Forum Discussions

Harder Integration

harder gradient

a little harder

Here's another (harder) one

I think I am making this harder

Search Tags