Is this problem right?












0












$begingroup$


Let $f$ be a continuous function from an open interval $Isubseteqmathbb{R}$ to a Banach space $E$. The problem/exercise asks to prove that $f$ is differentiable in $x_0in I$ if and only if the limit
$$lim_{(h,k)to (0,0)}frac{f(x_0+h)-f(x_0-h)}{h+k}$$
exists when $h,kgt 0$.



Shouldn't it be
$$frac{f(x_0+h)-f(x_0-k)}{h+k}$$
otherwise i don't see what is $k$ doing there in the denominator. I've been able to do the exercise asuming it wanted to say "$k$" there because, if not, then $0=lim_{kto0^+}(lim_{hto0^+})$ can be very different to $lim_{hto0^+}(lim_{kto0^+})$ and they should be equal. Should I try it as it is?? The book this is from doesn't mess around and the $h$ could be on purpose to make the exercise very difficult.



Thanks






limits derivatives







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  • 1




    $begingroup$
    Surely, there is a typo in the exercise.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:24
















0












$begingroup$


Let $f$ be a continuous function from an open interval $Isubseteqmathbb{R}$ to a Banach space $E$. The problem/exercise asks to prove that $f$ is differentiable in $x_0in I$ if and only if the limit
$$lim_{(h,k)to (0,0)}frac{f(x_0+h)-f(x_0-h)}{h+k}$$
exists when $h,kgt 0$.



Shouldn't it be
$$frac{f(x_0+h)-f(x_0-k)}{h+k}$$
otherwise i don't see what is $k$ doing there in the denominator. I've been able to do the exercise asuming it wanted to say "$k$" there because, if not, then $0=lim_{kto0^+}(lim_{hto0^+})$ can be very different to $lim_{hto0^+}(lim_{kto0^+})$ and they should be equal. Should I try it as it is?? The book this is from doesn't mess around and the $h$ could be on purpose to make the exercise very difficult.



Thanks






limits derivatives







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Surely, there is a typo in the exercise.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:24














0












0








0





$begingroup$


Let $f$ be a continuous function from an open interval $Isubseteqmathbb{R}$ to a Banach space $E$. The problem/exercise asks to prove that $f$ is differentiable in $x_0in I$ if and only if the limit
$$lim_{(h,k)to (0,0)}frac{f(x_0+h)-f(x_0-h)}{h+k}$$
exists when $h,kgt 0$.



Shouldn't it be
$$frac{f(x_0+h)-f(x_0-k)}{h+k}$$
otherwise i don't see what is $k$ doing there in the denominator. I've been able to do the exercise asuming it wanted to say "$k$" there because, if not, then $0=lim_{kto0^+}(lim_{hto0^+})$ can be very different to $lim_{hto0^+}(lim_{kto0^+})$ and they should be equal. Should I try it as it is?? The book this is from doesn't mess around and the $h$ could be on purpose to make the exercise very difficult.



Thanks






limits derivatives







share|cite|improve this question









$endgroup$




Let $f$ be a continuous function from an open interval $Isubseteqmathbb{R}$ to a Banach space $E$. The problem/exercise asks to prove that $f$ is differentiable in $x_0in I$ if and only if the limit
$$lim_{(h,k)to (0,0)}frac{f(x_0+h)-f(x_0-h)}{h+k}$$
exists when $h,kgt 0$.



Shouldn't it be
$$frac{f(x_0+h)-f(x_0-k)}{h+k}$$
otherwise i don't see what is $k$ doing there in the denominator. I've been able to do the exercise asuming it wanted to say "$k$" there because, if not, then $0=lim_{kto0^+}(lim_{hto0^+})$ can be very different to $lim_{hto0^+}(lim_{kto0^+})$ and they should be equal. Should I try it as it is?? The book this is from doesn't mess around and the $h$ could be on purpose to make the exercise very difficult.



Thanks






limits derivatives




limits derivatives






share|cite|improve this question













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asked Jan 3 at 23:21









PedroPedro

517212




517212








  • 1




    $begingroup$
    Surely, there is a typo in the exercise.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:24














  • 1




    $begingroup$
    Surely, there is a typo in the exercise.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:24








1




1




$begingroup$
Surely, there is a typo in the exercise.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:24




$begingroup$
Surely, there is a typo in the exercise.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:24










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You are right. It is obviously a typo.






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    $begingroup$

    You are right. It is obviously a typo.






    share|cite|improve this answer









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      1












      $begingroup$

      You are right. It is obviously a typo.






      share|cite|improve this answer









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        $begingroup$

        You are right. It is obviously a typo.






        share|cite|improve this answer









        $endgroup$



        You are right. It is obviously a typo.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Jan 3 at 23:24









        José Carlos SantosJosé Carlos Santos

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