Mixing
Existence of solutions of the equation with a limit.
$begingroup$
Let f be continuous function on [0,1] and
$$lim_{x→0} frac{f(x + frac13) + f(x + frac23)}{x}=1$$
Prove that exist $x_{0}in[0,1]$ which satisfies equation $f(x_{0})=0$
I suppouse that the numerator should approach $0$ which would implicate that for
x near $0$ $f(x + frac13)$ would be of opposite sign then $f(x + frac23)$ or both be $0$.
Then by intermediate value theorem we would know that there is $x_{0}in[frac13,frac23]$ which fulfill $f(x_{0})=0$
Yet, I have no idea how to prove that numerator $rightarrow 0$
limits continuity
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
$endgroup$
add a comment |
$begingroup$
Let f be continuous function on [0,1] and
$$lim_{x→0} frac{f(x + frac13) + f(x + frac23)}{x}=1$$
Prove that exist $x_{0}in[0,1]$ which satisfies equation $f(x_{0})=0$
I suppouse that the numerator should approach $0$ which would implicate that for
x near $0$ $f(x + frac13)$ would be of opposite sign then $f(x + frac23)$ or both be $0$.
Then by intermediate value theorem we would know that there is $x_{0}in[frac13,frac23]$ which fulfill $f(x_{0})=0$
Yet, I have no idea how to prove that numerator $rightarrow 0$
limits continuity
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
$endgroup$
add a comment |
$begingroup$
Let f be continuous function on [0,1] and
$$lim_{x→0} frac{f(x + frac13) + f(x + frac23)}{x}=1$$
Prove that exist $x_{0}in[0,1]$ which satisfies equation $f(x_{0})=0$
I suppouse that the numerator should approach $0$ which would implicate that for
x near $0$ $f(x + frac13)$ would be of opposite sign then $f(x + frac23)$ or both be $0$.
Then by intermediate value theorem we would know that there is $x_{0}in[frac13,frac23]$ which fulfill $f(x_{0})=0$
Yet, I have no idea how to prove that numerator $rightarrow 0$
limits continuity
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
$endgroup$
Let f be continuous function on [0,1] and
$$lim_{x→0} frac{f(x + frac13) + f(x + frac23)}{x}=1$$
Prove that exist $x_{0}in[0,1]$ which satisfies equation $f(x_{0})=0$
I suppouse that the numerator should approach $0$ which would implicate that for
x near $0$ $f(x + frac13)$ would be of opposite sign then $f(x + frac23)$ or both be $0$.
Then by intermediate value theorem we would know that there is $x_{0}in[frac13,frac23]$ which fulfill $f(x_{0})=0$
Yet, I have no idea how to prove that numerator $rightarrow 0$
limits continuity
limits continuity
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
asked Aug 4 '15 at 8:45
dyrAnddyrAnd
16511
16511
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Clearly we can see that $$lim_{x to 0}f(x + 1/3) + f(x + 2/3) = lim_{x to 0}xcdotfrac{f(x + 1/3) + f(x + 2/3)}{x} = 0 cdot 1 = 0$$ and by continuity of $f$ we see that this implies $f(1/3) + f(2/3) = 0$. If $f(1/3) = 0$ then we are done. If $f(1/3) neq 0$ then both $f(1/3)$ and $f(2/3)$ are of opposite signs and hence by intermediate value theorem there is an $x_{0} in (1/3, 2/3)$ for which $f(x_{0}) = 0$.
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
$endgroup$
add a comment |
$begingroup$
Hint: If the numerator doesn't go to $0$, can the limit goes to 1? At first suppose that there exists $lim_{xto 0} (f(x+1/3)+f(x+2/3))$ and isn't zero.
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
$endgroup$
add a comment |
$begingroup$
First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$lim_{xto 0} f(x+frac13) + f(x+frac23)neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(frac23) + f(frac13)=0$, and then use a well known theorem to finish your proof.
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
$endgroup$
add a comment |
$begingroup$
If
$$
lim frac{f(x)}{g(x)}=0
$$
and
$$
lim g(x)=0,
$$
then
$$
lim f(x) = lim frac{f(x)}{g(x)} cdot g(x) =0.
$$
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
$endgroup$
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
add a comment |
$begingroup$
I have a slightly different idea:
The codintion indicates that: $f(x+1/3) + f(x+2/3)$ is an equivalent infinitesimal to the function $x$.., which means that $f(x+1/3)+f(x+2/3)$ is equivalent to $ o(x)$, when $x$ is infinitely small...
So $f(x+1/3)+f(x+2/3)approx a_0 + a_1x$ and $a_0 = 0$, $a_1 = 0$ or $1$.
$a_1=0$ apparently leads to the result..
if $a_1 = 1$, it $f(x+1/3)+f(1+2/3)=x$. so $f(1/3)+f(2/3)=0$, which also leads to the conclusion.
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Clearly we can see that $$lim_{x to 0}f(x + 1/3) + f(x + 2/3) = lim_{x to 0}xcdotfrac{f(x + 1/3) + f(x + 2/3)}{x} = 0 cdot 1 = 0$$ and by continuity of $f$ we see that this implies $f(1/3) + f(2/3) = 0$. If $f(1/3) = 0$ then we are done. If $f(1/3) neq 0$ then both $f(1/3)$ and $f(2/3)$ are of opposite signs and hence by intermediate value theorem there is an $x_{0} in (1/3, 2/3)$ for which $f(x_{0}) = 0$.
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
$endgroup$
add a comment |
$begingroup$
Clearly we can see that $$lim_{x to 0}f(x + 1/3) + f(x + 2/3) = lim_{x to 0}xcdotfrac{f(x + 1/3) + f(x + 2/3)}{x} = 0 cdot 1 = 0$$ and by continuity of $f$ we see that this implies $f(1/3) + f(2/3) = 0$. If $f(1/3) = 0$ then we are done. If $f(1/3) neq 0$ then both $f(1/3)$ and $f(2/3)$ are of opposite signs and hence by intermediate value theorem there is an $x_{0} in (1/3, 2/3)$ for which $f(x_{0}) = 0$.
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
$endgroup$
add a comment |
$begingroup$
Clearly we can see that $$lim_{x to 0}f(x + 1/3) + f(x + 2/3) = lim_{x to 0}xcdotfrac{f(x + 1/3) + f(x + 2/3)}{x} = 0 cdot 1 = 0$$ and by continuity of $f$ we see that this implies $f(1/3) + f(2/3) = 0$. If $f(1/3) = 0$ then we are done. If $f(1/3) neq 0$ then both $f(1/3)$ and $f(2/3)$ are of opposite signs and hence by intermediate value theorem there is an $x_{0} in (1/3, 2/3)$ for which $f(x_{0}) = 0$.
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
$endgroup$
Clearly we can see that $$lim_{x to 0}f(x + 1/3) + f(x + 2/3) = lim_{x to 0}xcdotfrac{f(x + 1/3) + f(x + 2/3)}{x} = 0 cdot 1 = 0$$ and by continuity of $f$ we see that this implies $f(1/3) + f(2/3) = 0$. If $f(1/3) = 0$ then we are done. If $f(1/3) neq 0$ then both $f(1/3)$ and $f(2/3)$ are of opposite signs and hence by intermediate value theorem there is an $x_{0} in (1/3, 2/3)$ for which $f(x_{0}) = 0$.
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
answered Aug 4 '15 at 9:21
Paramanand SinghParamanand Singh
51.2k557169
51.2k557169
add a comment |
add a comment |
$begingroup$
Hint: If the numerator doesn't go to $0$, can the limit goes to 1? At first suppose that there exists $lim_{xto 0} (f(x+1/3)+f(x+2/3))$ and isn't zero.
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
$endgroup$
add a comment |
$begingroup$
Hint: If the numerator doesn't go to $0$, can the limit goes to 1? At first suppose that there exists $lim_{xto 0} (f(x+1/3)+f(x+2/3))$ and isn't zero.
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
$endgroup$
add a comment |
$begingroup$
Hint: If the numerator doesn't go to $0$, can the limit goes to 1? At first suppose that there exists $lim_{xto 0} (f(x+1/3)+f(x+2/3))$ and isn't zero.
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
$endgroup$
Hint: If the numerator doesn't go to $0$, can the limit goes to 1? At first suppose that there exists $lim_{xto 0} (f(x+1/3)+f(x+2/3))$ and isn't zero.
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
answered Aug 4 '15 at 8:49
Euler88 ...Euler88 ...
1,961513
1,961513
add a comment |
add a comment |
$begingroup$
First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$lim_{xto 0} f(x+frac13) + f(x+frac23)neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(frac23) + f(frac13)=0$, and then use a well known theorem to finish your proof.
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
$endgroup$
add a comment |
$begingroup$
First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$lim_{xto 0} f(x+frac13) + f(x+frac23)neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(frac23) + f(frac13)=0$, and then use a well known theorem to finish your proof.
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
$endgroup$
add a comment |
$begingroup$
First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$lim_{xto 0} f(x+frac13) + f(x+frac23)neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(frac23) + f(frac13)=0$, and then use a well known theorem to finish your proof.
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
$endgroup$
First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$lim_{xto 0} f(x+frac13) + f(x+frac23)neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(frac23) + f(frac13)=0$, and then use a well known theorem to finish your proof.
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
answered Aug 4 '15 at 8:54
5xum5xum
91.9k394161
91.9k394161
add a comment |
add a comment |
$begingroup$
If
$$
lim frac{f(x)}{g(x)}=0
$$
and
$$
lim g(x)=0,
$$
then
$$
lim f(x) = lim frac{f(x)}{g(x)} cdot g(x) =0.
$$
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
$endgroup$
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
add a comment |
$begingroup$
If
$$
lim frac{f(x)}{g(x)}=0
$$
and
$$
lim g(x)=0,
$$
then
$$
lim f(x) = lim frac{f(x)}{g(x)} cdot g(x) =0.
$$
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
$endgroup$
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
add a comment |
$begingroup$
If
$$
lim frac{f(x)}{g(x)}=0
$$
and
$$
lim g(x)=0,
$$
then
$$
lim f(x) = lim frac{f(x)}{g(x)} cdot g(x) =0.
$$
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
$endgroup$
If
$$
lim frac{f(x)}{g(x)}=0
$$
and
$$
lim g(x)=0,
$$
then
$$
lim f(x) = lim frac{f(x)}{g(x)} cdot g(x) =0.
$$
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
answered Aug 4 '15 at 8:53
SiminoreSiminore
30.5k33569
30.5k33569
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
add a comment |
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
$begingroup$
I think I do not understand. What $f$ and $g$ should I choose in this example?
$endgroup$
– dyrAnd
Aug 4 '15 at 8:55
add a comment |
$begingroup$
I have a slightly different idea:
The codintion indicates that: $f(x+1/3) + f(x+2/3)$ is an equivalent infinitesimal to the function $x$.., which means that $f(x+1/3)+f(x+2/3)$ is equivalent to $ o(x)$, when $x$ is infinitely small...
So $f(x+1/3)+f(x+2/3)approx a_0 + a_1x$ and $a_0 = 0$, $a_1 = 0$ or $1$.
$a_1=0$ apparently leads to the result..
if $a_1 = 1$, it $f(x+1/3)+f(1+2/3)=x$. so $f(1/3)+f(2/3)=0$, which also leads to the conclusion.
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
$endgroup$
add a comment |
$begingroup$
I have a slightly different idea:
The codintion indicates that: $f(x+1/3) + f(x+2/3)$ is an equivalent infinitesimal to the function $x$.., which means that $f(x+1/3)+f(x+2/3)$ is equivalent to $ o(x)$, when $x$ is infinitely small...
So $f(x+1/3)+f(x+2/3)approx a_0 + a_1x$ and $a_0 = 0$, $a_1 = 0$ or $1$.
$a_1=0$ apparently leads to the result..
if $a_1 = 1$, it $f(x+1/3)+f(1+2/3)=x$. so $f(1/3)+f(2/3)=0$, which also leads to the conclusion.
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
$endgroup$
add a comment |
$begingroup$
I have a slightly different idea:
The codintion indicates that: $f(x+1/3) + f(x+2/3)$ is an equivalent infinitesimal to the function $x$.., which means that $f(x+1/3)+f(x+2/3)$ is equivalent to $ o(x)$, when $x$ is infinitely small...
So $f(x+1/3)+f(x+2/3)approx a_0 + a_1x$ and $a_0 = 0$, $a_1 = 0$ or $1$.
$a_1=0$ apparently leads to the result..
if $a_1 = 1$, it $f(x+1/3)+f(1+2/3)=x$. so $f(1/3)+f(2/3)=0$, which also leads to the conclusion.
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
$endgroup$
I have a slightly different idea:
The codintion indicates that: $f(x+1/3) + f(x+2/3)$ is an equivalent infinitesimal to the function $x$.., which means that $f(x+1/3)+f(x+2/3)$ is equivalent to $ o(x)$, when $x$ is infinitely small...
So $f(x+1/3)+f(x+2/3)approx a_0 + a_1x$ and $a_0 = 0$, $a_1 = 0$ or $1$.
$a_1=0$ apparently leads to the result..
if $a_1 = 1$, it $f(x+1/3)+f(1+2/3)=x$. so $f(1/3)+f(2/3)=0$, which also leads to the conclusion.
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
answered Aug 4 '15 at 9:36
guangleiguanglei
352315
352315
add a comment |
add a comment |
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