Mixing
Is this problem right?
$begingroup$
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
asked Jan 3 at 23:21
PedroPedro
517212
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 23:21
PedroPedro
517212
$endgroup$
1
$begingroup$
Surely, there is a typo in the exercise.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:24
add a comment |
$begingroup$
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
asked Jan 3 at 23:21
PedroPedro
517212
$endgroup$
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
asked Jan 3 at 23:21
PedroPedro
517212
asked Jan 3 at 23:21
PedroPedro
517212
asked Jan 3 at 23:21
PedroPedro
517212
asked Jan 3 at 23:21
PedroPedro
517212
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right. It is obviously a typo.
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
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$begingroup$
You are right. It is obviously a typo.
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
You are right. It is obviously a typo.
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
You are right. It is obviously a typo.
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
You are right. It is obviously a typo.
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 3 at 23:24
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
add a comment |
add a comment |
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$begingroup$
Surely, there is a typo in the exercise.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:24