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

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)/h,
h -> 0, Analytic->True]


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

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

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


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)/h, h -> 0, Assumptions -> a + b == 1,
Analytic -> True]

• Your first conclusion is wrong: I would expect f - a f[h]-b f[2 h]== o[h] which leads to (1-a-b) f==0 Aug 21 '20 at 8:33
• @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. Aug 21 '20 at 8:47