L'Hospital's Rule

[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.

[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

[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.

[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.