How can I prove that no limit exists for this function over this particular interval?
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I was given the following function: $$ f(x) = begin{cases} frac{1}{q} &text{if }x = frac{p}{q} text{ is rational, reduced to lowest terms} \ 0 &text{if }x text{ is irrational}end{cases}$$
And was asked to attempt to prove or disprove whether or not, for some $j$ on the interval $(0,1)$, $lim limits_{x to j}{}f(x)$ exists.
Intuitively, I know that I can find an irrational number between any two rationals which would obviously result in some discontinuity. I considered the idea that maybe it could be possible assuming that among the infinitely many rationals on the interval which also share a common denominator and could be used to find the limit, but that obviously would not work. However, I cannot formulate any proper line of reason to prove the limit does not exist.
How can I prove this, as formally as possible?
limits irrational-numbers rational-numbers
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
$endgroup$
add a comment |
$begingroup$
I was given the following function: $$ f(x) = begin{cases} frac{1}{q} &text{if }x = frac{p}{q} text{ is rational, reduced to lowest terms} \ 0 &text{if }x text{ is irrational}end{cases}$$
And was asked to attempt to prove or disprove whether or not, for some $j$ on the interval $(0,1)$, $lim limits_{x to j}{}f(x)$ exists.
Intuitively, I know that I can find an irrational number between any two rationals which would obviously result in some discontinuity. I considered the idea that maybe it could be possible assuming that among the infinitely many rationals on the interval which also share a common denominator and could be used to find the limit, but that obviously would not work. However, I cannot formulate any proper line of reason to prove the limit does not exist.
How can I prove this, as formally as possible?
limits irrational-numbers rational-numbers
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
$endgroup$
2
$begingroup$
This called Thomae's function, see here en.m.wikipedia.org/wiki/Thomae%27s_function. The limit exists at irrational points.
$endgroup$
– Tom.
Jan 24 at 4:32
add a comment |
$begingroup$
I was given the following function: $$ f(x) = begin{cases} frac{1}{q} &text{if }x = frac{p}{q} text{ is rational, reduced to lowest terms} \ 0 &text{if }x text{ is irrational}end{cases}$$
And was asked to attempt to prove or disprove whether or not, for some $j$ on the interval $(0,1)$, $lim limits_{x to j}{}f(x)$ exists.
Intuitively, I know that I can find an irrational number between any two rationals which would obviously result in some discontinuity. I considered the idea that maybe it could be possible assuming that among the infinitely many rationals on the interval which also share a common denominator and could be used to find the limit, but that obviously would not work. However, I cannot formulate any proper line of reason to prove the limit does not exist.
How can I prove this, as formally as possible?
limits irrational-numbers rational-numbers
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
$endgroup$
I was given the following function: $$ f(x) = begin{cases} frac{1}{q} &text{if }x = frac{p}{q} text{ is rational, reduced to lowest terms} \ 0 &text{if }x text{ is irrational}end{cases}$$
And was asked to attempt to prove or disprove whether or not, for some $j$ on the interval $(0,1)$, $lim limits_{x to j}{}f(x)$ exists.
Intuitively, I know that I can find an irrational number between any two rationals which would obviously result in some discontinuity. I considered the idea that maybe it could be possible assuming that among the infinitely many rationals on the interval which also share a common denominator and could be used to find the limit, but that obviously would not work. However, I cannot formulate any proper line of reason to prove the limit does not exist.
How can I prove this, as formally as possible?
limits irrational-numbers rational-numbers
limits irrational-numbers rational-numbers
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
asked Jan 24 at 4:23
nine-hundrednine-hundred
30011
30011
2
$begingroup$
This called Thomae's function, see here en.m.wikipedia.org/wiki/Thomae%27s_function. The limit exists at irrational points.
$endgroup$
– Tom.
Jan 24 at 4:32
add a comment |
2
$begingroup$
This called Thomae's function, see here en.m.wikipedia.org/wiki/Thomae%27s_function. The limit exists at irrational points.
$endgroup$
– Tom.
Jan 24 at 4:32
2
2
$begingroup$
This called Thomae's function, see here en.m.wikipedia.org/wiki/Thomae%27s_function. The limit exists at irrational points.
$endgroup$
– Tom.
Jan 24 at 4:32
$begingroup$
This called Thomae's function, see here en.m.wikipedia.org/wiki/Thomae%27s_function. The limit exists at irrational points.
$endgroup$
– Tom.
Jan 24 at 4:32
add a comment |
1 Answer
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Recall the definition of $lim_{x to j}f(x)=a$:
$$mbox{Given} varepsilon > 0 mbox{there exists} delta > 0\ mbox{such that} left| f(x) - aright|<varepsilon\ mbox{whenever} 0<left| x-j right| <delta$$
Now for any $j$, given $varepsilon>0$, if we choose $delta$ small enough we can ensure that the denominator $q$ of any fraction $frac{p}{q}$ (in its lowest terms) in the interval $left(j-delta,j+deltaright)$ satisfies $q>frac{1}{varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0leleft| f(x)right|<varepsilon$ whenever $0<left|x-jright|<delta$.
Plugging this into the definition of a limit we see:
$$lim_{xto j}f(x) = 0 mbox{for all} jinleft(0,1right)$$
Now a function $g$ is continuous at $j$ iff $lim_{xto j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
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add a comment |
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$begingroup$
Recall the definition of $lim_{x to j}f(x)=a$:
$$mbox{Given} varepsilon > 0 mbox{there exists} delta > 0\ mbox{such that} left| f(x) - aright|<varepsilon\ mbox{whenever} 0<left| x-j right| <delta$$
Now for any $j$, given $varepsilon>0$, if we choose $delta$ small enough we can ensure that the denominator $q$ of any fraction $frac{p}{q}$ (in its lowest terms) in the interval $left(j-delta,j+deltaright)$ satisfies $q>frac{1}{varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0leleft| f(x)right|<varepsilon$ whenever $0<left|x-jright|<delta$.
Plugging this into the definition of a limit we see:
$$lim_{xto j}f(x) = 0 mbox{for all} jinleft(0,1right)$$
Now a function $g$ is continuous at $j$ iff $lim_{xto j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
$endgroup$
add a comment |
$begingroup$
Recall the definition of $lim_{x to j}f(x)=a$:
$$mbox{Given} varepsilon > 0 mbox{there exists} delta > 0\ mbox{such that} left| f(x) - aright|<varepsilon\ mbox{whenever} 0<left| x-j right| <delta$$
Now for any $j$, given $varepsilon>0$, if we choose $delta$ small enough we can ensure that the denominator $q$ of any fraction $frac{p}{q}$ (in its lowest terms) in the interval $left(j-delta,j+deltaright)$ satisfies $q>frac{1}{varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0leleft| f(x)right|<varepsilon$ whenever $0<left|x-jright|<delta$.
Plugging this into the definition of a limit we see:
$$lim_{xto j}f(x) = 0 mbox{for all} jinleft(0,1right)$$
Now a function $g$ is continuous at $j$ iff $lim_{xto j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
$endgroup$
add a comment |
$begingroup$
Recall the definition of $lim_{x to j}f(x)=a$:
$$mbox{Given} varepsilon > 0 mbox{there exists} delta > 0\ mbox{such that} left| f(x) - aright|<varepsilon\ mbox{whenever} 0<left| x-j right| <delta$$
Now for any $j$, given $varepsilon>0$, if we choose $delta$ small enough we can ensure that the denominator $q$ of any fraction $frac{p}{q}$ (in its lowest terms) in the interval $left(j-delta,j+deltaright)$ satisfies $q>frac{1}{varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0leleft| f(x)right|<varepsilon$ whenever $0<left|x-jright|<delta$.
Plugging this into the definition of a limit we see:
$$lim_{xto j}f(x) = 0 mbox{for all} jinleft(0,1right)$$
Now a function $g$ is continuous at $j$ iff $lim_{xto j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
$endgroup$
Recall the definition of $lim_{x to j}f(x)=a$:
$$mbox{Given} varepsilon > 0 mbox{there exists} delta > 0\ mbox{such that} left| f(x) - aright|<varepsilon\ mbox{whenever} 0<left| x-j right| <delta$$
Now for any $j$, given $varepsilon>0$, if we choose $delta$ small enough we can ensure that the denominator $q$ of any fraction $frac{p}{q}$ (in its lowest terms) in the interval $left(j-delta,j+deltaright)$ satisfies $q>frac{1}{varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0leleft| f(x)right|<varepsilon$ whenever $0<left|x-jright|<delta$.
Plugging this into the definition of a limit we see:
$$lim_{xto j}f(x) = 0 mbox{for all} jinleft(0,1right)$$
Now a function $g$ is continuous at $j$ iff $lim_{xto j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
edited Jan 24 at 8:02
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
answered Jan 24 at 7:51
Bernard HurleyBernard Hurley
1787
1787
add a comment |
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$begingroup$
This called Thomae's function, see here en.m.wikipedia.org/wiki/Thomae%27s_function. The limit exists at irrational points.
$endgroup$
– Tom.
Jan 24 at 4:32