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This question is provoked by an (in fact standard) solution to one of Baby Rudin’s exercises.

Suppose $f$ is defined in a neighborhood of $x$, and suppose $f^{\prime\prime}(x)$ >exists. Show that $$ \lim_{h \to 0} \frac{ f(x+h)+f(x-h)-2f(x)}{h^2} = >f^{\prime\prime}(x).$$ Hint: Use Theorem 5.13. [i.e. L'Hospital's rule]

What confuses me is the first step, using L'Hospital's rule:

\begin{align} \lim_{h \to 0} \frac{f(x+h) + f(x-h) -2f(x) }{h^2} &= \lim_{h \to 0} \frac{f^\prime(x+h) - f^\prime(x-h)}{2h} \\ & \ \ \ \mbox{ [ using L'Hospital's rule, } \\ & \ \ \ \ \ \ \mbox{ because $f(x+h)+f(x-h)-2f(x) \to 0$ and $h \to 0$ } \\ & \ \ \ \ \ \ \ \ \ \mbox{ as $h \to 0$ ]} \\ &= \lim_{h \to 0} \frac{ f^\prime(x+h) - f^\prime(x) + f^\prime(x) - f^\prime(x-h) }{2h} \end{align}

etc. But I believe the rule is about equating $\tfrac{f(x)}{g(x)}$ and $\tfrac{f^\prime(x)}{g^\prime(x)}$, i.e. ratio of functions resp. derivatives w.r.t the same variable, but were $g(x)=h^2$, we would have $g^\prime(x)=0$ for all $x \in \rm{Dom} \it f$, thus destroying the ratio. So are we permitted to differentiate w.r.t $x$ in the numerator and w.r.t $h$ in the denominator?

Any help is appreciated.

C_Arietta_C
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  • I don't understand. In the usual presentation, there's only $x$. If you had, e.g., $\lim_{x\to \infty}\frac {f(x)}{h^2}$, that would just be $\frac 1{h^2}\times \lim_{x\to \infty} f(x)$. In the second presentation, there is only $h$. – lulu Dec 12 '23 at 12:33
  • @lulu That’s exactly what’s giving me trouble, apparently in the second presentation we are taking an ‘alien’ derivative in the denominator: $(h^2)^\prime=2h$. – C_Arietta_C Dec 12 '23 at 12:55
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    You are differentitating with respect to $h$ in both numerator and denominator. $x$ is fixed throughout. I agree that writing $f'$ is a bit confusing since normally that would mean $\frac d{dx}f$. But note the sign change. $\frac d{dx}f(x-h)$ would just be $f'(x-h)$ while $\frac d{dh} f(x-h)=-f'(x-h)$, where of course I am perpetuating the abuse of notation, in both instances, $f'$ refers to the appropriate derivative. – lulu Dec 12 '23 at 12:57
  • @lulu But if we are diffentiating uniformly, i.e. with respect to $h$, will the resulting $f^\prime(x+h)$ (i.e. with notation abuse) be the same? The proof is completed by evoking the definition of $f^{\prime\prime}(x)$ as $\lim_{h \to 0} \frac{ f^\prime(x+h) - f^\prime(x) }{h} $ etc., where $f^\prime(x+h)$ resumes its common denotation. – C_Arietta_C Dec 13 '23 at 00:50
  • I suggest ditching that notation, it causes nothing but confusion. Just use $\frac d{dh}$, or even better $\frac {\partial}{\partial h}$ and so on. But, sure. Those functions are the same, as functions. Consider $f(x)=x^2$ as an example. Then $F(x,h)=f(x+h)=x^2+2hx+h^2$ and so$\frac {\partial}{\partial h}F(x,h)=2h+2x$ which is also $\frac {\partial}{\partial x}F(x,h)$. I'm sure this equality is why the author is comfortable with the notation, but I still think it is confusing. And, of course, that equality requires proof...which I assume he supplies? – lulu Dec 13 '23 at 01:07
  • To be sure, you need only remark that nothing is changed under a transposition under $x,h$ but it still merits being pointed out. – lulu Dec 13 '23 at 01:09
  • @lulu No, no proof is given for the «transposition invariance» in the solution, but I find it surprisingly simple, let$u=x+h,; g(x, h)=f(u),; $then$\frac{\partial g}{\partial h}=\frac{\partial g}{\partial u} \frac{\partial u}{\partial h}=\frac{\partial g}{\partial u}=\frac{\partial g}{\partial u}\frac{\partial u}{\partial x}=\frac{\partial g}{\partial x}$, anyway the invariance may as well be explained by symmetry. – C_Arietta_C Dec 13 '23 at 05:33
  • @lulu What I overlooked seems to be the fact that $x$ and $h$ are parameters with respect to each other, hence $\frac{\partial g}{\partial h}$ is itself a function about $x$ and vice versa, its value therefore still depending on the latter. – C_Arietta_C Dec 13 '23 at 05:33
  • I agree, it's not difficult. But I'd say the discussion the author provides is at least a little confusing. – lulu Dec 13 '23 at 09:18

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