$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$
$begingroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
$endgroup$
add a comment |
$begingroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
$endgroup$
add a comment |
$begingroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
$endgroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
355
add a comment |
add a comment |
2 Answers
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$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,396412
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
$endgroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7241419
1,7241419
add a comment |
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,396412
$endgroup$
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,396412
$endgroup$
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,396412
$endgroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,396412
answered Dec 18 '18 at 22:27
MikeMike
4,396412
answered Dec 18 '18 at 22:27
MikeMike
4,396412
answered Dec 18 '18 at 22:27
MikeMike
4,396412
4,396412
add a comment |
add a comment |
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