Formula derivation of the Second derivative

Question

I'm trying to prove the below equation:

The given function $f$ is twice differentiable. And the given formula is

$f''(x) = 2\lim_{h \rightarrow 0} \frac{f(x+h) -f(x) - f'(x)h } { h^2}$

I know this can be proved by l'Hospital's rule.

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Since the numerator and denominator goes to zero as $h \rightarrow 0$, the given formula becomes $lim_{h \rightarrow 0} \frac{f'(x+h)-f'(x)} {h}$, which equals to $f''(x)$.

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But the reason I write this post is this:

When I tried to change the $f'(x)$ of the numerator, by definition of the derivative, the equation no longer true.

The trial is

$f''(x)$ = $2\lim_{h \rightarrow 0} \frac{f(x+h) -f(x) - f'(x)h } { h^2}$ = $2\lim_{h \rightarrow 0} \frac{f(x+h) -f(x) - h\cdot lim_{h \rightarrow 0} \frac{f(x) - f(x-h)} {h} } { h^2}$ = $2\lim_{h \rightarrow 0} \frac{f(x+h) -2f(x) + f(x-h) } { h^2}$ = $2f''(x)$.

But $f''(x)$ $=$ $2f''(x)$ can't be true for $f''(x)$ isn't zero. The plausible inference for the obtaining the false result is the "change of $f'(x)$ to limit", from the second to the third of the above equalities, is wrong.

I think the possible explanation is, when the derivative change to the limit, the limit variable doesn't necessarily be $h$. The choice of the limit variable doesn't affect the derivative of the $f$, for example, $f'(x) = \lim_{k \rightarrow 0} \frac{f(x+k) -f(x) } {k}$ is also okay. So the formula can't be combined by $lim_{h \rightarrow 0}$.

Any help would be appreciated.

Answer 1

Note that Taylor's theorem tells us that $$ f(x + h) = f(x) + f'(x) h + \frac 12 f''(x) h^2 + o(h^2). $$ Using a $k$ for the inner limit provides us with some insight. In particular, we have $$ \lim_{h \to 0}\frac{f(x + h) - f(x) - f'(x) h}{h^2} = \lim_{h \to 0}\frac{f(x + h) - f(x) - h\lim_{k \to 0}\frac{f(x) - f(x-k)}{k} }{h^2}\\ = \lim_{h \to 0}\lim_{k \to 0} \frac{f(x+h) - f(x) - \frac hk f(x) + \frac hk f(x-k)}{h^2}. $$ Here, $\lim_{h \to 0}\lim_{k \to 0}$ indicates that we take a limit first with respect to $k$, then with respect to $h$.

Note that taking $k$ to be a function of $h$ and computing a limit over $h$ produces a different result. For your example, we see that taking $k(h) = h$ yields a limit of $2f''(x)$. More generally, taking $k(h) = h/a$ for $a \neq 0$ yields $$ \lim_{h \to 0} \frac{f(x+h) - f(x) - \frac h{k(h)} f(x) + \frac h{k(h)} f(x-k(h))}{h^2}\\ = \lim_{h \to 0} \frac{f(x+h) - f(x) - af(x) + a f(x-h/a)}{h^2}\\ = \lim_{h \to 0} \frac{f(x+h) - (1 + a)f(x)+ a f(x-h/a)}{h^2}\\ = \lim_{h \to 0} \frac 1{h^2}[f(x) + hf'(x) + \frac 12f''(x)h^2 - (1 + a)f(x)\\ \qquad + a (f(x) - f'(x) (h/a) + \frac 12 f''(x) (h/a)^2) + o(h^2)]\\ = \lim_{h \to 0} \frac{\frac 12(1 + 1/a)f''(x)h^2 + o(h^2)}{h^2} \\ = \frac{a + 1}{2a} f''(x). $$