function defined by the average rate of change Given a differentiable function $f(x)$, let $g(x)$ be defined by
$g(x) = \begin{cases} (f(x)-f(a))/(x-a) &\mbox{if } x \neq a \\ 
f'(a) & \mbox{if } x = a. \end{cases}$
Suppose also f(x) is twice differentiable, then I guess that $g'(a)=f''(a)/2$ since $$g'(a)= \lim \frac{g(x)-g(a)}{x-a}=\lim \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2}$$ and applying L'Hospital's rule twice. But how can I prove this without using L'Hospital's rule?
 A: Here's a more direct approach.   To simplify the writing I let a = 0. 
If f''(x) exists, then  f(x) = f(0) + f'(0)x + f''(0)x2/2 + R(x) where limx→0R(x)/x = 0. 
f(x)-f(0) = f'(0)x + f''(x)x2/2 + R(x)   Since g(x) = [f(x)-f(0)]/x  we have g(x) = f'(0) + f''(0)x/2 + R(x)/x.  
Then g'(x) = f''(0)/2 + R(x)/x.  g'(0) = limx→0g'(x) = f''(0)/2, since R(x)/x goes to 0.
Yes, we are relying on the Taylor's expansion. 
A: Well, you can do it directly, altho it is more trouble.  Let F(x) = f(x) - f(a) - f'(a)(x-a) and G(x) = (x-a)2. Both F and G are differentiable.  Then by the definition for the derivative we know that
F(x)= F(a) + F'(a)(x-a) + R(x)  =   F'(a)(x-a) + R(x)  since F(a) = 0      

where limx→aR(x)/(x-a) = 0.  And similarly
G(x) = G(a) + G'(a))x-a) + S(x) = G'(a)(x-a) + S(x)  since G(a) = 0

same conditions on S(x).
So 
F(x)/G(x) = [F(x)/(x-a)[/[G(x)/(x-a)] = [F'(a) + R(x)/(x-a)]/[G'(a) + S(x)/(x-a)] 
Thus as per your computations
g'(a) = limx→a F(x)/G(x) = F'(a)/G'(a).  The R(x)/(x-a) and S(x)/(x-a) go to zero (that fact is buried in the definition of the derivative).
Okay, F'(a) = f''(a) and G'(a) = 2.  So you are done.
Wasn't L'Hospital's rule simpler?  Or more accurately, we have just given a proof of L'Hospital's rule for this particular case, which was more trouble than simply applying it.
Is there a more direct way to do this?  Maybe there is, but I didn't find it in the amount of time I had to spend on this.   
