Derivative definition with double limit












1














The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing







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  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50
















1














The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing







share|cite|improve this question






















  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50














1












1








1







The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing







share|cite|improve this question













The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing




real-analysis proof-verification proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 18:04









高田航

1,283318




1,283318












  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50


















  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50
















Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46






Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46














Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50




Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50










1 Answer
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$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer























  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13













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$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer























  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13


















2














$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer























  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13
















2












2








2






$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer














$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 15:15

























answered Nov 20 at 18:54









Federico

4,649514




4,649514












  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13




















  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13


















I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47




I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47




1




1




$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13






$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13




















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