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.
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*}