Find functions $f$ and $g$
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Find functions $f$ and $g$ defined on $(0,infty)$ such that $$lim_{xtoinfty} f(x)=infty$$ $$lim_{xtoinfty} g(x)=infty$$ and $$lim_{xtoinfty} (f-g)(x)=0$$.Can you find such fuch function with $g(x)gt 0$ for all $xin (0,infty)$ and $$lim_{xtoinfty} frac{f(x)}{g(x)}=0$$
Justification:I guess there do not exist such functions because for $$lim_{xtoinfty} (f-g)(x)=0$$ then it says that $$lim_{xtoinfty} f(x)=lim_{xtoinfty} g(x)=Lne 0$$ and if it so then $$lim_{xtoinfty} frac{f(x)}{g(x)}=1ne 0$$ which is contradiction.hence there do not exist such functions which satisfies given condition.
I am not sure about my justification.Is it correct?Please help me.
real-analysis limits
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
$endgroup$
add a comment |
$begingroup$
Find functions $f$ and $g$ defined on $(0,infty)$ such that $$lim_{xtoinfty} f(x)=infty$$ $$lim_{xtoinfty} g(x)=infty$$ and $$lim_{xtoinfty} (f-g)(x)=0$$.Can you find such fuch function with $g(x)gt 0$ for all $xin (0,infty)$ and $$lim_{xtoinfty} frac{f(x)}{g(x)}=0$$
Justification:I guess there do not exist such functions because for $$lim_{xtoinfty} (f-g)(x)=0$$ then it says that $$lim_{xtoinfty} f(x)=lim_{xtoinfty} g(x)=Lne 0$$ and if it so then $$lim_{xtoinfty} frac{f(x)}{g(x)}=1ne 0$$ which is contradiction.hence there do not exist such functions which satisfies given condition.
I am not sure about my justification.Is it correct?Please help me.
real-analysis limits
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
$endgroup$
$begingroup$
I believe your logic is okay, although I think there are some assumptions with assuming $frac{infty}{infty} = 1$ but I do not remember them.
$endgroup$
– Wolfy
Dec 21 '18 at 5:49
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I tried many examples but most of them are failed to satisfy all the conditions specially $$lim_{xtoinfty} (f-g)(x)=0$$ and $$lim_{xtoinfty} (f/g)(x)=0$$
$endgroup$
– ASHWINI SANKHE
Dec 21 '18 at 5:52
add a comment |
$begingroup$
Find functions $f$ and $g$ defined on $(0,infty)$ such that $$lim_{xtoinfty} f(x)=infty$$ $$lim_{xtoinfty} g(x)=infty$$ and $$lim_{xtoinfty} (f-g)(x)=0$$.Can you find such fuch function with $g(x)gt 0$ for all $xin (0,infty)$ and $$lim_{xtoinfty} frac{f(x)}{g(x)}=0$$
Justification:I guess there do not exist such functions because for $$lim_{xtoinfty} (f-g)(x)=0$$ then it says that $$lim_{xtoinfty} f(x)=lim_{xtoinfty} g(x)=Lne 0$$ and if it so then $$lim_{xtoinfty} frac{f(x)}{g(x)}=1ne 0$$ which is contradiction.hence there do not exist such functions which satisfies given condition.
I am not sure about my justification.Is it correct?Please help me.
real-analysis limits
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
$endgroup$
Find functions $f$ and $g$ defined on $(0,infty)$ such that $$lim_{xtoinfty} f(x)=infty$$ $$lim_{xtoinfty} g(x)=infty$$ and $$lim_{xtoinfty} (f-g)(x)=0$$.Can you find such fuch function with $g(x)gt 0$ for all $xin (0,infty)$ and $$lim_{xtoinfty} frac{f(x)}{g(x)}=0$$
Justification:I guess there do not exist such functions because for $$lim_{xtoinfty} (f-g)(x)=0$$ then it says that $$lim_{xtoinfty} f(x)=lim_{xtoinfty} g(x)=Lne 0$$ and if it so then $$lim_{xtoinfty} frac{f(x)}{g(x)}=1ne 0$$ which is contradiction.hence there do not exist such functions which satisfies given condition.
I am not sure about my justification.Is it correct?Please help me.
real-analysis limits
real-analysis limits
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
asked Dec 21 '18 at 5:43
ASHWINI SANKHEASHWINI SANKHE
11210
11210
$begingroup$
I believe your logic is okay, although I think there are some assumptions with assuming $frac{infty}{infty} = 1$ but I do not remember them.
$endgroup$
– Wolfy
Dec 21 '18 at 5:49
$begingroup$
I tried many examples but most of them are failed to satisfy all the conditions specially $$lim_{xtoinfty} (f-g)(x)=0$$ and $$lim_{xtoinfty} (f/g)(x)=0$$
$endgroup$
– ASHWINI SANKHE
Dec 21 '18 at 5:52
add a comment |
$begingroup$
I believe your logic is okay, although I think there are some assumptions with assuming $frac{infty}{infty} = 1$ but I do not remember them.
$endgroup$
– Wolfy
Dec 21 '18 at 5:49
$begingroup$
I tried many examples but most of them are failed to satisfy all the conditions specially $$lim_{xtoinfty} (f-g)(x)=0$$ and $$lim_{xtoinfty} (f/g)(x)=0$$
$endgroup$
– ASHWINI SANKHE
Dec 21 '18 at 5:52
$begingroup$
I believe your logic is okay, although I think there are some assumptions with assuming $frac{infty}{infty} = 1$ but I do not remember them.
$endgroup$
– Wolfy
Dec 21 '18 at 5:49
$begingroup$
I believe your logic is okay, although I think there are some assumptions with assuming $frac{infty}{infty} = 1$ but I do not remember them.
$endgroup$
– Wolfy
Dec 21 '18 at 5:49
$begingroup$
I tried many examples but most of them are failed to satisfy all the conditions specially $$lim_{xtoinfty} (f-g)(x)=0$$ and $$lim_{xtoinfty} (f/g)(x)=0$$
$endgroup$
– ASHWINI SANKHE
Dec 21 '18 at 5:52
$begingroup$
I tried many examples but most of them are failed to satisfy all the conditions specially $$lim_{xtoinfty} (f-g)(x)=0$$ and $$lim_{xtoinfty} (f/g)(x)=0$$
$endgroup$
– ASHWINI SANKHE
Dec 21 '18 at 5:52
add a comment |
1 Answer
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$displaystylelim_{xtoinfty}big(f(x)-g(x)big)=0$ and $displaystylelim_{xtoinfty}g(x)=infty$ imply $displaystylelim_{xtoinfty}frac{f(x)-g(x)}{g(x)}=0$, i.e. $displaystylelim_{xtoinfty}frac{f(x)}{g(x)}=1$.
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
$endgroup$
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1 Answer
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$begingroup$
$displaystylelim_{xtoinfty}big(f(x)-g(x)big)=0$ and $displaystylelim_{xtoinfty}g(x)=infty$ imply $displaystylelim_{xtoinfty}frac{f(x)-g(x)}{g(x)}=0$, i.e. $displaystylelim_{xtoinfty}frac{f(x)}{g(x)}=1$.
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
$endgroup$
add a comment |
$begingroup$
$displaystylelim_{xtoinfty}big(f(x)-g(x)big)=0$ and $displaystylelim_{xtoinfty}g(x)=infty$ imply $displaystylelim_{xtoinfty}frac{f(x)-g(x)}{g(x)}=0$, i.e. $displaystylelim_{xtoinfty}frac{f(x)}{g(x)}=1$.
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
$endgroup$
add a comment |
$begingroup$
$displaystylelim_{xtoinfty}big(f(x)-g(x)big)=0$ and $displaystylelim_{xtoinfty}g(x)=infty$ imply $displaystylelim_{xtoinfty}frac{f(x)-g(x)}{g(x)}=0$, i.e. $displaystylelim_{xtoinfty}frac{f(x)}{g(x)}=1$.
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
$endgroup$
$displaystylelim_{xtoinfty}big(f(x)-g(x)big)=0$ and $displaystylelim_{xtoinfty}g(x)=infty$ imply $displaystylelim_{xtoinfty}frac{f(x)-g(x)}{g(x)}=0$, i.e. $displaystylelim_{xtoinfty}frac{f(x)}{g(x)}=1$.
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
answered Dec 21 '18 at 6:06
metamorphymetamorphy
3,6921621
3,6921621
add a comment |
add a comment |
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$begingroup$
I believe your logic is okay, although I think there are some assumptions with assuming $frac{infty}{infty} = 1$ but I do not remember them.
$endgroup$
– Wolfy
Dec 21 '18 at 5:49
$begingroup$
I tried many examples but most of them are failed to satisfy all the conditions specially $$lim_{xtoinfty} (f-g)(x)=0$$ and $$lim_{xtoinfty} (f/g)(x)=0$$
$endgroup$
– ASHWINI SANKHE
Dec 21 '18 at 5:52