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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?

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  • $\begingroup$ Perhaps you can use mean value theorem, $g'(a)=\frac{g(x)-g(a)}{x-a}$? $\endgroup$ – Alex Aug 24 '13 at 2:43
  • $\begingroup$ @Alex What you write is not MVT. $\endgroup$ – Did Aug 26 '13 at 13:25
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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.

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  • $\begingroup$ Just a comment: How is replacing $a$ by $0$ "simplifying the writing"? It's not even shorter for you to type it. $\endgroup$ – TMM Aug 26 '13 at 16:54
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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.

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    $\begingroup$ It seems you rewrite the Taylor extension (as your previous answer) $\endgroup$ – RSh Aug 24 '13 at 5:19

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