Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$
up vote
5
down vote
favorite
Problem:
If $f(x)$ is continous at $x=0$, and $limlimits_{xto 0} dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|>1$, prove that $f'(0)$ exists and $f'(0)=dfrac{b}{a-1}$.
This approach is definitely wrong:
begin{align}
b&=lim_{xto 0} frac{f(ax)-f(x)}{x}\
&=lim_{xto 0} frac{f(ax)-f(0)-(f(x)-f(0))}{x}\
&=af'(0)-f'(0)\
&=(a-1)f'(0)
end{align}
I will show you a case why this approach is wrong:
[f(x)= begin{cases}
1,&xneq0\
0,&x=0
end{cases}]
$lim_{xto0}dfrac{f(3x)-f(x)}{x}=lim_{xto0} dfrac{1-1}{x}=0$
but $lim_{xto0}dfrac{f(3x)}{x}=infty$,$lim_{xto0}dfrac{f(x)}{x}=infty$
Does anyone know how to prove it? Thanks in advance!
calculus derivatives
add a comment |
up vote
5
down vote
favorite
Problem:
If $f(x)$ is continous at $x=0$, and $limlimits_{xto 0} dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|>1$, prove that $f'(0)$ exists and $f'(0)=dfrac{b}{a-1}$.
This approach is definitely wrong:
begin{align}
b&=lim_{xto 0} frac{f(ax)-f(x)}{x}\
&=lim_{xto 0} frac{f(ax)-f(0)-(f(x)-f(0))}{x}\
&=af'(0)-f'(0)\
&=(a-1)f'(0)
end{align}
I will show you a case why this approach is wrong:
[f(x)= begin{cases}
1,&xneq0\
0,&x=0
end{cases}]
$lim_{xto0}dfrac{f(3x)-f(x)}{x}=lim_{xto0} dfrac{1-1}{x}=0$
but $lim_{xto0}dfrac{f(3x)}{x}=infty$,$lim_{xto0}dfrac{f(x)}{x}=infty$
Does anyone know how to prove it? Thanks in advance!
calculus derivatives
Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses?
– Jonas Meyer
Jul 19 '16 at 8:41
the proof of the existence of $f'(0)$ is actually the problem is asking for
– Spaceship222
Jul 19 '16 at 8:47
Your result holds even if $|a| < 1$. See update to my answer.
– Paramanand Singh
Jul 19 '16 at 12:14
Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it?
– BusyAnt
Jul 19 '16 at 12:35
@BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$
– Spaceship222
Jul 19 '16 at 12:43
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Problem:
If $f(x)$ is continous at $x=0$, and $limlimits_{xto 0} dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|>1$, prove that $f'(0)$ exists and $f'(0)=dfrac{b}{a-1}$.
This approach is definitely wrong:
begin{align}
b&=lim_{xto 0} frac{f(ax)-f(x)}{x}\
&=lim_{xto 0} frac{f(ax)-f(0)-(f(x)-f(0))}{x}\
&=af'(0)-f'(0)\
&=(a-1)f'(0)
end{align}
I will show you a case why this approach is wrong:
[f(x)= begin{cases}
1,&xneq0\
0,&x=0
end{cases}]
$lim_{xto0}dfrac{f(3x)-f(x)}{x}=lim_{xto0} dfrac{1-1}{x}=0$
but $lim_{xto0}dfrac{f(3x)}{x}=infty$,$lim_{xto0}dfrac{f(x)}{x}=infty$
Does anyone know how to prove it? Thanks in advance!
calculus derivatives
Problem:
If $f(x)$ is continous at $x=0$, and $limlimits_{xto 0} dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|>1$, prove that $f'(0)$ exists and $f'(0)=dfrac{b}{a-1}$.
This approach is definitely wrong:
begin{align}
b&=lim_{xto 0} frac{f(ax)-f(x)}{x}\
&=lim_{xto 0} frac{f(ax)-f(0)-(f(x)-f(0))}{x}\
&=af'(0)-f'(0)\
&=(a-1)f'(0)
end{align}
I will show you a case why this approach is wrong:
[f(x)= begin{cases}
1,&xneq0\
0,&x=0
end{cases}]
$lim_{xto0}dfrac{f(3x)-f(x)}{x}=lim_{xto0} dfrac{1-1}{x}=0$
but $lim_{xto0}dfrac{f(3x)}{x}=infty$,$lim_{xto0}dfrac{f(x)}{x}=infty$
Does anyone know how to prove it? Thanks in advance!
calculus derivatives
calculus derivatives
edited Jul 19 '16 at 12:02
Paramanand Singh
48.4k555156
48.4k555156
asked Jul 19 '16 at 8:37
Spaceship222
5517
5517
Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses?
– Jonas Meyer
Jul 19 '16 at 8:41
the proof of the existence of $f'(0)$ is actually the problem is asking for
– Spaceship222
Jul 19 '16 at 8:47
Your result holds even if $|a| < 1$. See update to my answer.
– Paramanand Singh
Jul 19 '16 at 12:14
Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it?
– BusyAnt
Jul 19 '16 at 12:35
@BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$
– Spaceship222
Jul 19 '16 at 12:43
add a comment |
Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses?
– Jonas Meyer
Jul 19 '16 at 8:41
the proof of the existence of $f'(0)$ is actually the problem is asking for
– Spaceship222
Jul 19 '16 at 8:47
Your result holds even if $|a| < 1$. See update to my answer.
– Paramanand Singh
Jul 19 '16 at 12:14
Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it?
– BusyAnt
Jul 19 '16 at 12:35
@BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$
– Spaceship222
Jul 19 '16 at 12:43
Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses?
– Jonas Meyer
Jul 19 '16 at 8:41
Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses?
– Jonas Meyer
Jul 19 '16 at 8:41
the proof of the existence of $f'(0)$ is actually the problem is asking for
– Spaceship222
Jul 19 '16 at 8:47
the proof of the existence of $f'(0)$ is actually the problem is asking for
– Spaceship222
Jul 19 '16 at 8:47
Your result holds even if $|a| < 1$. See update to my answer.
– Paramanand Singh
Jul 19 '16 at 12:14
Your result holds even if $|a| < 1$. See update to my answer.
– Paramanand Singh
Jul 19 '16 at 12:14
Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it?
– BusyAnt
Jul 19 '16 at 12:35
Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it?
– BusyAnt
Jul 19 '16 at 12:35
@BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$
– Spaceship222
Jul 19 '16 at 12:43
@BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$
– Spaceship222
Jul 19 '16 at 12:43
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
This is a tricky question and the solution is somewhat non-obvious. We know that $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) to 0$ as $x to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, ldots, n$ we get $$f(x) - f(x/a^{n}) = bxsum_{k = 1}^{n}frac{1}{a^{k}} + xsum_{k = 1}^{n}frac{g(x/a^{k})}{a^{k}}$$ Letting $n to infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = frac{bx}{a - 1} + xsum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x to 0$ we get $$f'(0) = lim_{x to 0}frac{f(x) - f(0)}{x} = frac{b}{a - 1} + lim_{x to 0}sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$
The sum $$sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x to 0$ because $g(x) to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)to 0$ as $x to 0$, it follows that for any $epsilon > 0$ there is a $delta > 0$ such that $|g(x)| < epsilon$ for all $x$ with $0 <|x| < delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < delta$ if $|x| < delta$ and therefore $|g(x/a^{k})| < epsilon$. Thus if $0 < |x| < delta$ we have $$left|sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}right| < sum_{k = 1}^{infty}frac{epsilon}{|a|^{k}} = frac{epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x to 0$.
Hence $f'(0) = b/(a - 1)$.
BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ implies that $$lim_{t to 0}frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = frac{-bc}{c - 1} = frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| neq 1$.
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
1
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
|
show 4 more comments
up vote
0
down vote
In this quickly closed question the case $a=2$ is considered, which allows the following simpler solution:
Define $g(x):=f(x)- bx-f(0)$. Then $g$ is continuous at $0$, $g(0)=0$, and $$lim_{xto0}{g(2x)-g(x)over x}=0 .$$
We have to prove that $g'(0)=lim_{xto0}{g(x)over x}=0$.
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that $|g(2t)-g(t)|leqepsilon |t|$ for $0<tleqdelta$. Assume $|x|leqdelta$. Then for each $Nin{mathbb N}$ one has
$$g(x)=sum_{k=1}^Nbigl(g(x/2^{k-1})-g(x/2^k)bigr)+g(x/2^N) ,$$
and therefore
$$bigl|g(x)bigr|leqsum_{k=1}^Nepsilon,{|x|over 2^k} +g(x/2^N)leqepsilon|x|+g(x/2^N) .$$
Since $Nin{mathbb N}$ is arbitrary we in fact have $bigl|g(x)bigr|leq epsilon|x|$, or $left|{g(x)over x}right|leqepsilon$, and this for all $xin>]0,delta]$.
add a comment |
up vote
-1
down vote
Hint: You're very close.
Write the expression as $$afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}{x}$$ Note that $xto 0$ if and only if $axto 0$ (since $aneq 0$).
Can you see it from this?
2
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
1
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
1
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
1
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
2
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
|
show 12 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This is a tricky question and the solution is somewhat non-obvious. We know that $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) to 0$ as $x to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, ldots, n$ we get $$f(x) - f(x/a^{n}) = bxsum_{k = 1}^{n}frac{1}{a^{k}} + xsum_{k = 1}^{n}frac{g(x/a^{k})}{a^{k}}$$ Letting $n to infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = frac{bx}{a - 1} + xsum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x to 0$ we get $$f'(0) = lim_{x to 0}frac{f(x) - f(0)}{x} = frac{b}{a - 1} + lim_{x to 0}sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$
The sum $$sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x to 0$ because $g(x) to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)to 0$ as $x to 0$, it follows that for any $epsilon > 0$ there is a $delta > 0$ such that $|g(x)| < epsilon$ for all $x$ with $0 <|x| < delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < delta$ if $|x| < delta$ and therefore $|g(x/a^{k})| < epsilon$. Thus if $0 < |x| < delta$ we have $$left|sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}right| < sum_{k = 1}^{infty}frac{epsilon}{|a|^{k}} = frac{epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x to 0$.
Hence $f'(0) = b/(a - 1)$.
BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ implies that $$lim_{t to 0}frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = frac{-bc}{c - 1} = frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| neq 1$.
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
1
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
|
show 4 more comments
up vote
4
down vote
accepted
This is a tricky question and the solution is somewhat non-obvious. We know that $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) to 0$ as $x to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, ldots, n$ we get $$f(x) - f(x/a^{n}) = bxsum_{k = 1}^{n}frac{1}{a^{k}} + xsum_{k = 1}^{n}frac{g(x/a^{k})}{a^{k}}$$ Letting $n to infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = frac{bx}{a - 1} + xsum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x to 0$ we get $$f'(0) = lim_{x to 0}frac{f(x) - f(0)}{x} = frac{b}{a - 1} + lim_{x to 0}sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$
The sum $$sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x to 0$ because $g(x) to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)to 0$ as $x to 0$, it follows that for any $epsilon > 0$ there is a $delta > 0$ such that $|g(x)| < epsilon$ for all $x$ with $0 <|x| < delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < delta$ if $|x| < delta$ and therefore $|g(x/a^{k})| < epsilon$. Thus if $0 < |x| < delta$ we have $$left|sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}right| < sum_{k = 1}^{infty}frac{epsilon}{|a|^{k}} = frac{epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x to 0$.
Hence $f'(0) = b/(a - 1)$.
BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ implies that $$lim_{t to 0}frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = frac{-bc}{c - 1} = frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| neq 1$.
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
1
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
|
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This is a tricky question and the solution is somewhat non-obvious. We know that $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) to 0$ as $x to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, ldots, n$ we get $$f(x) - f(x/a^{n}) = bxsum_{k = 1}^{n}frac{1}{a^{k}} + xsum_{k = 1}^{n}frac{g(x/a^{k})}{a^{k}}$$ Letting $n to infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = frac{bx}{a - 1} + xsum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x to 0$ we get $$f'(0) = lim_{x to 0}frac{f(x) - f(0)}{x} = frac{b}{a - 1} + lim_{x to 0}sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$
The sum $$sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x to 0$ because $g(x) to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)to 0$ as $x to 0$, it follows that for any $epsilon > 0$ there is a $delta > 0$ such that $|g(x)| < epsilon$ for all $x$ with $0 <|x| < delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < delta$ if $|x| < delta$ and therefore $|g(x/a^{k})| < epsilon$. Thus if $0 < |x| < delta$ we have $$left|sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}right| < sum_{k = 1}^{infty}frac{epsilon}{|a|^{k}} = frac{epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x to 0$.
Hence $f'(0) = b/(a - 1)$.
BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ implies that $$lim_{t to 0}frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = frac{-bc}{c - 1} = frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| neq 1$.
This is a tricky question and the solution is somewhat non-obvious. We know that $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) to 0$ as $x to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, ldots, n$ we get $$f(x) - f(x/a^{n}) = bxsum_{k = 1}^{n}frac{1}{a^{k}} + xsum_{k = 1}^{n}frac{g(x/a^{k})}{a^{k}}$$ Letting $n to infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = frac{bx}{a - 1} + xsum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x to 0$ we get $$f'(0) = lim_{x to 0}frac{f(x) - f(0)}{x} = frac{b}{a - 1} + lim_{x to 0}sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$
The sum $$sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x to 0$ because $g(x) to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)to 0$ as $x to 0$, it follows that for any $epsilon > 0$ there is a $delta > 0$ such that $|g(x)| < epsilon$ for all $x$ with $0 <|x| < delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < delta$ if $|x| < delta$ and therefore $|g(x/a^{k})| < epsilon$. Thus if $0 < |x| < delta$ we have $$left|sum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}right| < sum_{k = 1}^{infty}frac{epsilon}{|a|^{k}} = frac{epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x to 0$.
Hence $f'(0) = b/(a - 1)$.
BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$lim_{x to 0}frac{f(ax) - f(x)}{x} = b$$ implies that $$lim_{t to 0}frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = frac{-bc}{c - 1} = frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| neq 1$.
edited Jul 25 '16 at 8:09
answered Jul 19 '16 at 9:18
Paramanand Singh
48.4k555156
48.4k555156
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
1
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
|
show 4 more comments
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
1
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
How "adding such equation" do you get the left side you did? Do you mean when adding over $;k;$ from $;1;$ to $;n;$? Then, you left $;ntoinfty;$ , but why would $;xto0implies;$ the second summand in the right tends to zero? Even if the right sum converges for all $;x>R;$ , for some $;R>0;$, how it depends on $;x;$ could makea difference.
– DonAntonio
Jul 19 '16 at 9:24
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:27
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
@DonAntonio: see my updated answer.
– Paramanand Singh
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied...
– DonAntonio
Jul 19 '16 at 9:32
1
1
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
@DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else.
– Paramanand Singh
Jul 19 '16 at 9:38
|
show 4 more comments
up vote
0
down vote
In this quickly closed question the case $a=2$ is considered, which allows the following simpler solution:
Define $g(x):=f(x)- bx-f(0)$. Then $g$ is continuous at $0$, $g(0)=0$, and $$lim_{xto0}{g(2x)-g(x)over x}=0 .$$
We have to prove that $g'(0)=lim_{xto0}{g(x)over x}=0$.
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that $|g(2t)-g(t)|leqepsilon |t|$ for $0<tleqdelta$. Assume $|x|leqdelta$. Then for each $Nin{mathbb N}$ one has
$$g(x)=sum_{k=1}^Nbigl(g(x/2^{k-1})-g(x/2^k)bigr)+g(x/2^N) ,$$
and therefore
$$bigl|g(x)bigr|leqsum_{k=1}^Nepsilon,{|x|over 2^k} +g(x/2^N)leqepsilon|x|+g(x/2^N) .$$
Since $Nin{mathbb N}$ is arbitrary we in fact have $bigl|g(x)bigr|leq epsilon|x|$, or $left|{g(x)over x}right|leqepsilon$, and this for all $xin>]0,delta]$.
add a comment |
up vote
0
down vote
In this quickly closed question the case $a=2$ is considered, which allows the following simpler solution:
Define $g(x):=f(x)- bx-f(0)$. Then $g$ is continuous at $0$, $g(0)=0$, and $$lim_{xto0}{g(2x)-g(x)over x}=0 .$$
We have to prove that $g'(0)=lim_{xto0}{g(x)over x}=0$.
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that $|g(2t)-g(t)|leqepsilon |t|$ for $0<tleqdelta$. Assume $|x|leqdelta$. Then for each $Nin{mathbb N}$ one has
$$g(x)=sum_{k=1}^Nbigl(g(x/2^{k-1})-g(x/2^k)bigr)+g(x/2^N) ,$$
and therefore
$$bigl|g(x)bigr|leqsum_{k=1}^Nepsilon,{|x|over 2^k} +g(x/2^N)leqepsilon|x|+g(x/2^N) .$$
Since $Nin{mathbb N}$ is arbitrary we in fact have $bigl|g(x)bigr|leq epsilon|x|$, or $left|{g(x)over x}right|leqepsilon$, and this for all $xin>]0,delta]$.
add a comment |
up vote
0
down vote
up vote
0
down vote
In this quickly closed question the case $a=2$ is considered, which allows the following simpler solution:
Define $g(x):=f(x)- bx-f(0)$. Then $g$ is continuous at $0$, $g(0)=0$, and $$lim_{xto0}{g(2x)-g(x)over x}=0 .$$
We have to prove that $g'(0)=lim_{xto0}{g(x)over x}=0$.
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that $|g(2t)-g(t)|leqepsilon |t|$ for $0<tleqdelta$. Assume $|x|leqdelta$. Then for each $Nin{mathbb N}$ one has
$$g(x)=sum_{k=1}^Nbigl(g(x/2^{k-1})-g(x/2^k)bigr)+g(x/2^N) ,$$
and therefore
$$bigl|g(x)bigr|leqsum_{k=1}^Nepsilon,{|x|over 2^k} +g(x/2^N)leqepsilon|x|+g(x/2^N) .$$
Since $Nin{mathbb N}$ is arbitrary we in fact have $bigl|g(x)bigr|leq epsilon|x|$, or $left|{g(x)over x}right|leqepsilon$, and this for all $xin>]0,delta]$.
In this quickly closed question the case $a=2$ is considered, which allows the following simpler solution:
Define $g(x):=f(x)- bx-f(0)$. Then $g$ is continuous at $0$, $g(0)=0$, and $$lim_{xto0}{g(2x)-g(x)over x}=0 .$$
We have to prove that $g'(0)=lim_{xto0}{g(x)over x}=0$.
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that $|g(2t)-g(t)|leqepsilon |t|$ for $0<tleqdelta$. Assume $|x|leqdelta$. Then for each $Nin{mathbb N}$ one has
$$g(x)=sum_{k=1}^Nbigl(g(x/2^{k-1})-g(x/2^k)bigr)+g(x/2^N) ,$$
and therefore
$$bigl|g(x)bigr|leqsum_{k=1}^Nepsilon,{|x|over 2^k} +g(x/2^N)leqepsilon|x|+g(x/2^N) .$$
Since $Nin{mathbb N}$ is arbitrary we in fact have $bigl|g(x)bigr|leq epsilon|x|$, or $left|{g(x)over x}right|leqepsilon$, and this for all $xin>]0,delta]$.
answered Nov 24 at 15:17
Christian Blatter
171k7111325
171k7111325
add a comment |
add a comment |
up vote
-1
down vote
Hint: You're very close.
Write the expression as $$afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}{x}$$ Note that $xto 0$ if and only if $axto 0$ (since $aneq 0$).
Can you see it from this?
2
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
1
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
1
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
1
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
2
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
|
show 12 more comments
up vote
-1
down vote
Hint: You're very close.
Write the expression as $$afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}{x}$$ Note that $xto 0$ if and only if $axto 0$ (since $aneq 0$).
Can you see it from this?
2
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
1
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
1
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
1
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
2
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
|
show 12 more comments
up vote
-1
down vote
up vote
-1
down vote
Hint: You're very close.
Write the expression as $$afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}{x}$$ Note that $xto 0$ if and only if $axto 0$ (since $aneq 0$).
Can you see it from this?
Hint: You're very close.
Write the expression as $$afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}{x}$$ Note that $xto 0$ if and only if $axto 0$ (since $aneq 0$).
Can you see it from this?
edited Jul 19 '16 at 12:31
answered Jul 19 '16 at 8:48
MPW
29.7k11856
29.7k11856
2
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
1
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
1
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
1
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
2
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
|
show 12 more comments
2
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
1
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
1
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
1
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
2
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
2
2
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
$lim_{xto0}frac{f(0)-f(x)}{x}$ is not the same as $lim_{xto0}frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$
– Spaceship222
Jul 19 '16 at 9:02
1
1
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
@MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one).
– DonAntonio
Jul 19 '16 at 9:03
1
1
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=lim_{xto0}frac{f(ax)-f(x)}x=lim_{xto0}left[afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}xright]=lim_{tto0}(a-1)frac{f(t)-t(0)}t$$because when $;xto0;$ both limits (without the constant $;a;$) within the parentheses are the same, whether it exists or not, because $;f;$ is given continuous at zero and thus it is the same to take $;lim f(x);$ or $;lim f(ax);$ when $xto0$. The rightmost expression, compared to the left side, answers all +1
– DonAntonio
Jul 19 '16 at 9:10
1
1
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect.
– Paramanand Singh
Jul 19 '16 at 9:20
2
2
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
@DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $lim_{x to 0^{+}}F(x)$ and $lim_{x to 0^{+}}F(ax)$ don't exist and yet $$lim_{x to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work.
– Paramanand Singh
Jul 19 '16 at 10:02
|
show 12 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1864058%2fprove-that-f0-exists-and-f0-b-a-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses?
– Jonas Meyer
Jul 19 '16 at 8:41
the proof of the existence of $f'(0)$ is actually the problem is asking for
– Spaceship222
Jul 19 '16 at 8:47
Your result holds even if $|a| < 1$. See update to my answer.
– Paramanand Singh
Jul 19 '16 at 12:14
Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it?
– BusyAnt
Jul 19 '16 at 12:35
@BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$
– Spaceship222
Jul 19 '16 at 12:43