How to use L'Hospital's rule for abstract functions

Question

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?

Answer 1

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

Try

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

Answer 2

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]