It is known that $f(x)$ has a first-order continuous derivative in a neighborhood of point x = 0 and $f(0) f^{\prime}(0) \neq 0$. When $h \rightarrow 0$,then $a f(h)+b f(2 h)-f(0)=o(h)$.I want to calculate the limit of the abstract function $\frac{a f(h)+b f(2 h)-f(0)}{h}$ by using L'Hospital's rule: $\lim _{h \rightarrow 0} \frac{a f(h)+b f(2 h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{a f^{\prime}(h)+2 b f^{\prime}(2 h)}{1}=(a+2b) f^{\prime}(0)$

Limit[(a*f[h] + b*f[2 h] - f[0])/h, 
 h -> 0, Analytic->True]

But the above methods can not get the desired results, what should I do to get the correct results?

  • 2
    $\begingroup$ Are you sure your maths is correct? I don't think the numerator limits to zero, in general. $\endgroup$
    – mikado
    Aug 21 '20 at 6:43
  • 1
    $\begingroup$ Try for instance with Limit[(a f[h] + h f[2 h] - a f[0])/h, h -> 0, Analytic -> True] which has the numerator $\to 0$ as h $\to 0$. $\endgroup$
    – Cesareo
    Aug 21 '20 at 7:28
  • $\begingroup$ @mikado Thank you for your guidance, I have updated the details of the question. $\endgroup$ Aug 21 '20 at 8:04
  • $\begingroup$ @Cesareo I have added the full details.. Thank you for your guidance. $\endgroup$ Aug 21 '20 at 8:10

L'Hospital's rule is applicable if the Limit would give a singular expression 0/0


Limit[(a*f[h] + b*f[2 h] - (a + b) f[0])/h, h -> 0, Analytic -> True]
(*(a + 2 b) Derivative[1][f][0]*)    

Only if $ a f(h)+b f(2 h)-f(0)$ is an infinitive that tends to zero, we can use the L'Hospital's rule.

Because when $h \rightarrow 0$,then $a f(h)+b f(2 h)-f(0)=o(h)$, so $(1-a-b) f(0)=0$. And because $f(0) f^{\prime}(0) \neq 0$, $a+b=1$.

Limit[(a f[h] + b f[2 h] - f[0])/h, h -> 0, Assumptions -> a + b == 1,
  Analytic -> True]
  • 1
    $\begingroup$ Your first conclusion is wrong: I would expect f[0] - a f[h]-b f[2 h]== o[h] which leads to (1-a-b) f[0]==0 $\endgroup$ Aug 21 '20 at 8:33
  • $\begingroup$ @UlrichNeumann Yes, because $f(0) f^{\prime}(0) \neq 0$, $f(0) \neq 0$ and $(1-a-b)=0$. I need to think about it again. $\endgroup$ Aug 21 '20 at 8:47

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