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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.

Lorenzo B.
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amir na
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  • Have you tried Taylor expansion? – gammatester Nov 24 '18 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. – amir na Nov 24 '18 at 18:37
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    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. – Paramanand Singh Nov 25 '18 at 01:09
  • @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. – user21820 Sep 12 '21 at 13:25

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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*}

xpaul
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