Finding Limit related to third derivative without L'Hospital

Let $$f(x)$$ be a real function and we know that $$f'(x) , f''(x) , f'''(x)$$ are defined on the domain of the function.

Find $$\lim_{h\to 0} \frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}$$

I can find this limit using L'Hospital's rule and it is equal to $$8f'''(x)$$ But I want to know how can I find the limit without using L'Hospital's rule.

Also using more advanced methods like Taylor series, etc. are not valid. Only obvious things like definition of derivative $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ and some equation like $$a=a+b-b$$ are valid.

• Have you tried Taylor expansion? Commented Nov 24, 2018 at 18:22
• @gammatester The question is meant to be solved without using any advanced method. Only the definition of derivative and some obvious equations like $a = a + b -b$ are valid. I edited the question for more clarification. Commented Nov 24, 2018 at 18:37
• A solution to this problem requires the use of L'Hospital's Rule or Taylor or an argument which is equivalent to the proof of these theorems. It is not a matter of simple algebraic manipulation and then application of algebra of limits. Commented Nov 25, 2018 at 1:09
• @ParamanandSingh is correct, and any claim that it can be done without essentially proving asymptotic expansion up to the required order is simply bogus. If you are not allowed to use Taylor series or L'Hopital, then the best way is simply to prove the needed asymptotic expansion, which is actually quite easy; see here. Commented Sep 12, 2021 at 13:25

By the Taylor Theorem, write $$f(x+h)$$ as $$f(x+h)=f(x)+f'(x)h+\frac12f''(x)h^2+\frac16f'''(x)h^3+R(x,h)h^3$$ where the Peano remainder $$R(x,h)$$ satisfies $$\lim_{h\to0}R(x,h)=0.$$ Then $$f(x+h)-f(x-h)=2f'(x)h+\frac13f'''(x)h^3+(R(x,h)-R(x,-h))h^3$$ and $$f(x+3h)-f(x-3h)=6f'(x)h+9f'''(x)h^3+27(R(x,3h)-R(x,-3h))h^3.$$ Therefore $$\begin{eqnarray*} &&\lim_{h\to 0} \frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}\\ &=&\lim_{h\to 0} \frac{\bigg[f(x+3h)- f(x-3h)\bigg] - 3\bigg[ f(x+h)- f(x-h) \bigg]}{h^3}\\ &=&\lim_{h\to 0} \frac{8f'''(x)h^3+27\bigg[R(x,3h)-R(x,-3h)\bigg]h^3-3\bigg[R(x,h)-R(x,-h)\bigg]h^3}{h^3}\\ &=&8f'''(x)+\lim_{h\to0}\bigg\{27\bigg[R(x,3h)-R(x,-3h)\bigg]-3\bigg[R(x,h)-R(x,-h)\bigg]\bigg\}\\ &=&f'''(x). \end{eqnarray*}$$