1. ## help with two limit problems please!

1) Limit as x-->1- (from the right) of: sqrt(1-x^2) / sqrt(1-x^3)

(I know that I need to apply L'Hopital's rule here, maybe even twice, but I keep getting the wrong answer. Thanks in advance if you can help!)

2) Let f be a twice differentiable function and fix a value of x.
(a) Show that
lim as h-->0: [f(x+h)-f(x-h)]/2h = f '(x)
(b) Show that
lim as h-->0: [f(x+h)-2f(x)+f(x-h)]/(h^2) = f ''(x)

2. Hello,

(a) Show that
lim as h-->0: [f(x+h)-f(x-h)]/2h = f '(x)
You know that $\boxed{\frac{f(x+h)-f(x)}{h}=f'(x)}$ (lim h to 0)

If you substitute h for -h, you have :

$\frac{f(x-h)-f(x)}{-h}=f'(x) \Longleftrightarrow \boxed{\frac{-f(x-h)+f(x)}{h}=f'(x)}$ (lim h to 0)

Now, add the two boxed equations term by term

$2f'(x)=\dots$

1) Limit as x-->1- (from the right) of: sqrt(1-x^2) / sqrt(1-x^3)

(I know that I need to apply L'Hopital's rule here, maybe even twice, but I keep getting the wrong answer. Thanks in advance if you can help!)
The best to do is to show what you've done, it will be easier to point out your mistakes