# L'Hospital's Rule

1. Jul 26, 2004

### Moose352

Having a very hard time solving this problem:

Use L'Hospital's rule to show that

$$\lim_{h\rightarrow0}\frac{f(x+h)-f(x-h)}{2h}=f^{'}(x)$$

Thanks.

Last edited: Jul 26, 2004
2. Jul 26, 2004

### hello3719

I can prove it without using l'hopital rule

we know that f'(x) =lim as h->0 (f(x+h) - f(x) / h) by definition
and also f'(x) = lim as h-> 0 (f(x) - f(x-h) /h )

if we add both of them we get 2(f'(x)) = (f(x+h) - f(x) + f(x) - f(x-h) ) / h

2(f'(x)) = (f(x+h) - f(x-h) ) / h
divide by 2 on both sides
f'(x) = (f(x+h) - f(x-h) ) / 2h

3. Jul 26, 2004

### Moose352

Thanks for the solution. In fact, I did the same thing, except in the other way. But the signs were not working for some dumb reason, dumb stuff kept cancelling out.

4. Jul 27, 2004

### Galileo

Just use the chain rule:
$$\frac{d}{dh}f(x\pm h)=\pm f'(x \pm h)$$
so you get
$$\frac{f'(x+h)+f'(x-h)}{2}$$
then take the limit as h goes to zero.