Finding Limit related to third derivative without L'Hospital












1














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 $$lim_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}$$



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 $$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$ and some equation like $a=a+b-b$ are valid.






calculus limits limits-without-lhopital







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  • Have you tried Taylor expansion?
    – gammatester
    Nov 24 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 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.
    – Paramanand Singh
    Nov 25 at 1:09


















1














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 $$lim_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}$$



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 $$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$ and some equation like $a=a+b-b$ are valid.






calculus limits limits-without-lhopital







share|cite|improve this question
























  • Have you tried Taylor expansion?
    – gammatester
    Nov 24 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 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.
    – Paramanand Singh
    Nov 25 at 1:09
















1












1








1







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 $$lim_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}$$



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 $$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$ and some equation like $a=a+b-b$ are valid.






calculus limits limits-without-lhopital







share|cite|improve this question















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 $$lim_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}$$



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 $$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$ and some equation like $a=a+b-b$ are valid.






calculus limits limits-without-lhopital




calculus limits limits-without-lhopital






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edited Nov 27 at 20:02









Lorenzo B.

1,8222520




1,8222520










asked Nov 24 at 18:01









amir na

385




385












  • Have you tried Taylor expansion?
    – gammatester
    Nov 24 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 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.
    – Paramanand Singh
    Nov 25 at 1:09




















  • Have you tried Taylor expansion?
    – gammatester
    Nov 24 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 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.
    – Paramanand Singh
    Nov 25 at 1:09


















Have you tried Taylor expansion?
– gammatester
Nov 24 at 18:22




Have you tried Taylor expansion?
– gammatester
Nov 24 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 at 18:37




@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 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.
– Paramanand Singh
Nov 25 at 1:09






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 at 1:09












2 Answers
2






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oldest

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2














Using the definition of a derivative, in symmetric form, as
$$f'(x) = lim_{h to 0} frac{f(x+h) - f(x-h)}{2 , h}$$
then the second derivative follows
begin{align}
f''(x) &= lim_{h to 0} frac{f'(x+h) - f'(x-h)}{2 , h} \
&= lim_{h to 0} frac{f(x+ 2h) - 2 f(x) + f(x- 2h)}{4 , h^2}.
end{align}
Taking another derivative yields
begin{align}
f'''(x) &= lim_{h to 0} frac{f'(x+ 2h) - 2 f'(x) + f'(x- 2h)}{4 , h^2} \
&= lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{8 , h^3}.
end{align}



The last result can be restated as
$$8 , f'''(x) = lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{h^3}.$$






share|cite|improve this answer

















  • 2




    Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
    – Rolf Hoyer
    Nov 24 at 19:29












  • Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
    – Paramanand Singh
    Nov 25 at 1:07





















1














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
$$ lim_{hto0}R(x,h)=0. $$
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_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}\
&=&lim_{hto 0} frac{bigg[f(x+3h)- f(x-3h)bigg] - 3bigg[ f(x+h)- f(x-h) bigg]}{h^3}\
&=&lim_{hto 0} frac{8f'''(x)h^3+27bigg[R(x,3h)-R(x,-3h)bigg]h^3-3bigg[R(x,h)-R(x,-h)bigg]h^3}{h^3}\
&=&8f'''(x)+lim_{hto0}bigg{27bigg[R(x,3h)-R(x,-3h)bigg]-3bigg[R(x,h)-R(x,-h)bigg]bigg}\
&=&f'''(x).
end{eqnarray*}






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    2 Answers
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    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Using the definition of a derivative, in symmetric form, as
    $$f'(x) = lim_{h to 0} frac{f(x+h) - f(x-h)}{2 , h}$$
    then the second derivative follows
    begin{align}
    f''(x) &= lim_{h to 0} frac{f'(x+h) - f'(x-h)}{2 , h} \
    &= lim_{h to 0} frac{f(x+ 2h) - 2 f(x) + f(x- 2h)}{4 , h^2}.
    end{align}
    Taking another derivative yields
    begin{align}
    f'''(x) &= lim_{h to 0} frac{f'(x+ 2h) - 2 f'(x) + f'(x- 2h)}{4 , h^2} \
    &= lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{8 , h^3}.
    end{align}



    The last result can be restated as
    $$8 , f'''(x) = lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{h^3}.$$






    share|cite|improve this answer

















    • 2




      Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
      – Rolf Hoyer
      Nov 24 at 19:29












    • Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
      – Paramanand Singh
      Nov 25 at 1:07


















    2














    Using the definition of a derivative, in symmetric form, as
    $$f'(x) = lim_{h to 0} frac{f(x+h) - f(x-h)}{2 , h}$$
    then the second derivative follows
    begin{align}
    f''(x) &= lim_{h to 0} frac{f'(x+h) - f'(x-h)}{2 , h} \
    &= lim_{h to 0} frac{f(x+ 2h) - 2 f(x) + f(x- 2h)}{4 , h^2}.
    end{align}
    Taking another derivative yields
    begin{align}
    f'''(x) &= lim_{h to 0} frac{f'(x+ 2h) - 2 f'(x) + f'(x- 2h)}{4 , h^2} \
    &= lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{8 , h^3}.
    end{align}



    The last result can be restated as
    $$8 , f'''(x) = lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{h^3}.$$






    share|cite|improve this answer

















    • 2




      Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
      – Rolf Hoyer
      Nov 24 at 19:29












    • Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
      – Paramanand Singh
      Nov 25 at 1:07
















    2












    2








    2






    Using the definition of a derivative, in symmetric form, as
    $$f'(x) = lim_{h to 0} frac{f(x+h) - f(x-h)}{2 , h}$$
    then the second derivative follows
    begin{align}
    f''(x) &= lim_{h to 0} frac{f'(x+h) - f'(x-h)}{2 , h} \
    &= lim_{h to 0} frac{f(x+ 2h) - 2 f(x) + f(x- 2h)}{4 , h^2}.
    end{align}
    Taking another derivative yields
    begin{align}
    f'''(x) &= lim_{h to 0} frac{f'(x+ 2h) - 2 f'(x) + f'(x- 2h)}{4 , h^2} \
    &= lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{8 , h^3}.
    end{align}



    The last result can be restated as
    $$8 , f'''(x) = lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{h^3}.$$






    share|cite|improve this answer












    Using the definition of a derivative, in symmetric form, as
    $$f'(x) = lim_{h to 0} frac{f(x+h) - f(x-h)}{2 , h}$$
    then the second derivative follows
    begin{align}
    f''(x) &= lim_{h to 0} frac{f'(x+h) - f'(x-h)}{2 , h} \
    &= lim_{h to 0} frac{f(x+ 2h) - 2 f(x) + f(x- 2h)}{4 , h^2}.
    end{align}
    Taking another derivative yields
    begin{align}
    f'''(x) &= lim_{h to 0} frac{f'(x+ 2h) - 2 f'(x) + f'(x- 2h)}{4 , h^2} \
    &= lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{8 , h^3}.
    end{align}



    The last result can be restated as
    $$8 , f'''(x) = lim_{h to 0} frac{f(x+ 3h) - 3 f(x+h) + 3 f(x-h) - f(x- 3h)}{h^3}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 18:46









    Leucippus

    19.6k102871




    19.6k102871








    • 2




      Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
      – Rolf Hoyer
      Nov 24 at 19:29












    • Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
      – Paramanand Singh
      Nov 25 at 1:07
















    • 2




      Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
      – Rolf Hoyer
      Nov 24 at 19:29












    • Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
      – Paramanand Singh
      Nov 25 at 1:07










    2




    2




    Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
    – Rolf Hoyer
    Nov 24 at 19:29






    Replacing a $f'(x+h)$ inside a limit as $hto 0$ with $ frac{f(x+2h)-f(x)}{h}$ feels like it's obviously valid, but when you look at it carefully you're making an implicit simplification along the lines of $$lim_{hto 0}left(lim_{kto 0} frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}right) = lim_{hto 0} frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which seems less obvious to me. For this particular case the proofs justifying this interchange rely on continuity of $f''$, which is fine, but I'm worried when you do a similar interchange later it implicitly relies on $f'''$ being continuous
    – Rolf Hoyer
    Nov 24 at 19:29














    Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
    – Paramanand Singh
    Nov 25 at 1:07






    Your first and second use of equal sign is fine, but further usage is not automatic/obvious and needs to be justified using L'Hospital's Rule or Taylor.
    – Paramanand Singh
    Nov 25 at 1:07













    1














    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
    $$ lim_{hto0}R(x,h)=0. $$
    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_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}\
    &=&lim_{hto 0} frac{bigg[f(x+3h)- f(x-3h)bigg] - 3bigg[ f(x+h)- f(x-h) bigg]}{h^3}\
    &=&lim_{hto 0} frac{8f'''(x)h^3+27bigg[R(x,3h)-R(x,-3h)bigg]h^3-3bigg[R(x,h)-R(x,-h)bigg]h^3}{h^3}\
    &=&8f'''(x)+lim_{hto0}bigg{27bigg[R(x,3h)-R(x,-3h)bigg]-3bigg[R(x,h)-R(x,-h)bigg]bigg}\
    &=&f'''(x).
    end{eqnarray*}






    share|cite|improve this answer




























      1














      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
      $$ lim_{hto0}R(x,h)=0. $$
      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_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}\
      &=&lim_{hto 0} frac{bigg[f(x+3h)- f(x-3h)bigg] - 3bigg[ f(x+h)- f(x-h) bigg]}{h^3}\
      &=&lim_{hto 0} frac{8f'''(x)h^3+27bigg[R(x,3h)-R(x,-3h)bigg]h^3-3bigg[R(x,h)-R(x,-h)bigg]h^3}{h^3}\
      &=&8f'''(x)+lim_{hto0}bigg{27bigg[R(x,3h)-R(x,-3h)bigg]-3bigg[R(x,h)-R(x,-h)bigg]bigg}\
      &=&f'''(x).
      end{eqnarray*}






      share|cite|improve this answer


























        1












        1








        1






        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
        $$ lim_{hto0}R(x,h)=0. $$
        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_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}\
        &=&lim_{hto 0} frac{bigg[f(x+3h)- f(x-3h)bigg] - 3bigg[ f(x+h)- f(x-h) bigg]}{h^3}\
        &=&lim_{hto 0} frac{8f'''(x)h^3+27bigg[R(x,3h)-R(x,-3h)bigg]h^3-3bigg[R(x,h)-R(x,-h)bigg]h^3}{h^3}\
        &=&8f'''(x)+lim_{hto0}bigg{27bigg[R(x,3h)-R(x,-3h)bigg]-3bigg[R(x,h)-R(x,-h)bigg]bigg}\
        &=&f'''(x).
        end{eqnarray*}






        share|cite|improve this answer














        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
        $$ lim_{hto0}R(x,h)=0. $$
        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_{hto 0} frac{f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)}{h^3}\
        &=&lim_{hto 0} frac{bigg[f(x+3h)- f(x-3h)bigg] - 3bigg[ f(x+h)- f(x-h) bigg]}{h^3}\
        &=&lim_{hto 0} frac{8f'''(x)h^3+27bigg[R(x,3h)-R(x,-3h)bigg]h^3-3bigg[R(x,h)-R(x,-h)bigg]h^3}{h^3}\
        &=&8f'''(x)+lim_{hto0}bigg{27bigg[R(x,3h)-R(x,-3h)bigg]-3bigg[R(x,h)-R(x,-h)bigg]bigg}\
        &=&f'''(x).
        end{eqnarray*}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 27 at 19:41

























        answered Nov 26 at 21:56









        xpaul

        22.4k14455




        22.4k14455






























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