Using the same limit for a second derivative

Solution 1:

Simple answer: You are right that by definition of derivative the expression you gave for the second derivative is the correct one. However it turns out that the other one is equal, though not obviously.

Proof by L'Hopital

One way to prove it is to use L'Hopital's rule once to get:

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

because numerator and denominator are zero when $h = 0$ and are differentiable with respect to $h$. What we get is not quite the definition of the second derivative so we have to manipulate:

$\lim_{h \to 0} \dfrac{f'(x+2h)-f'(x+h)}{h} = \lim_{h \to 0} 2\dfrac{f'(x+2h)-f'(x)}{2h} - \lim_{h \to 0} \dfrac{f'(x+h)-f'(x)}{h}$

which is valid because both limits on the right exist. Note that:

$\lim_{h \to 0} 2\dfrac{f'(x+2h)-f'(x)}{2h} = 2 \lim_{h \to 0} \dfrac{f'(x+h)-f'(x)}{h}$

since $h \to 0$ is equivalent to $2h \to 0$ or just going by the definition of limit, and the $2$ factors out of the limit. Finally we get:

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

by definition of second derivative.

Note

Notice that $f$ must be differentiable in an open interval around $x$ which was critical for L'Hopital's rule to work. But $f'$ need not be differentiable or even continuous in an open interval around $x$. All that is necessary, as used in the proof, is that $f'$ is differentiable at the single point $x$.

Solution 2:

Another completely different method and arguably more intuitive is to use asymptotic expansions.

Proof by asymptotic expansion

For any differentiable function $f$ such that $f'$ is differentiable at $x$ we have:

$f(x+h) \in f(x) + f'(x) h + \frac{1}{2}f''(x) h^2 + o(h^2)$ as $h \to 0$

This uses Little-O-notation so you may want to look at that if you've not come across it. First let us use it to solve the question:

As $h \to 0$:

  $f(x+2h) \in f(x) + f'(x) 2h + \frac{1}{2} f''(x) 4h^2 + o(h^2)$

  $f(x+h) \in f(x) + f'(x) h + \frac{1}{2} f''(x) h^2 + o(h^2)$

  $f(x+2h)-2f(x+h)+f(x) \in f''(x) h^2 + o(h^2)$

  $\dfrac{f(x+2h)-2f(x+h)+f(x)}{h^2} \in f''(x) + o(1)$

Therefore $\dfrac{f(x+2h)-2f(x+h)+f(x)}{h^2} \to f''(x)$ as $h \to 0$.

Proof of the asymptotic expansion $\def\rr{\mathbb{R}}$

Let $g(h) = f(x) + f'(x) h + \frac{1}{2}f''(x) h^2$ for any $h \in \rr$.

Then $g'(h) = f'(x) + f''(x) h$ for any $h \in \rr$.

Also $f'(x+h) \in f'(x) + f''(x) h + o(h)$ as $h \to 0$ [by definition of derivative of $f'$].

Thus $f'(x+h) - g'(h) \in o(h)$ as $h \to 0$.

$f(x+h) - g(h) = ( f'(x+c) - g'(c) ) h$ for some $c \in [0,h]$ [by the mean value theorem]

$\ \in o(h) h = o(h^2)$ as $h \to 0$.

Notes

The above proof easily extends to higher derivatives, and it is easy to prove any similar expressions for them.