Proving an Intuitive Result RigorouslyProve that $sumlimits_{r=1}^n (sr^{-1} -r^{-3}) -s log n$ has a limit...
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Proving an Intuitive Result Rigorously
Prove that $sumlimits_{r=1}^n (sr^{-1} -r^{-3}) -s log n$ has a limit as $n to infty$, $forall s>1$Did I take this limit correctly?$lim_{krightarrow infty}frac{2^k}{gamma}logmathbb{E}[e^{-gamma frac{X}{2^k}}]$Why “$limlimits_{xrightarrow infty} frac{x+sin x}{x}$ does not exist” is not an acceptable answer?Proving that $lim_{x {to} infty}f(x)=infty $Is $n^2=O(2^{log(n)})$?Finding $limsup_{nrightarrowinfty} n^{frac{log(n)}{n}}$Rate of growth of linear and logarithmic functionsEvaluate $lim_{xrightarrow0^+}{-xlog{x}}$Proving Limit Rigorously
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
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I have found that for distinct functions (not constants) $f(x),g(x),h(x)$ that this is true:
$$lim_{x to infty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=1$$ I want to know: why is that, ie is there a rigorous proof for why that is? Maybe a paper to turn to or is it really simple?
EDIT: I realized I probably should have put my actual problem first. I hope this gives more context. For this limit:
$$lim_{nrightarrowinfty}frac{e^{H(n)}log H(n)}{e^gamma nlog(gamma+log n)+frac{n}{loglog n}}$$
$H(n)$ tends to $e^gamma n$, which is the same as $gamma + log n$. So, you can reword it as this:
$$lim_{nrightarrowinfty}frac{e^{gamma + log n}log(gamma + log n)}{(gamma + log n)log(gamma+log n)+frac{n}{loglog n}}$$
So how do you continue to prove that this limit is $1$?
calculus limits
asked 10 hours ago
Quote DaveQuote Dave
12111
$endgroup$
|
show 7 more comments
$begingroup$
I have found that for distinct functions (not constants) $f(x),g(x),h(x)$ that this is true:
$$lim_{x to infty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=1$$ I want to know: why is that, ie is there a rigorous proof for why that is? Maybe a paper to turn to or is it really simple?
EDIT: I realized I probably should have put my actual problem first. I hope this gives more context. For this limit:
$$lim_{nrightarrowinfty}frac{e^{H(n)}log H(n)}{e^gamma nlog(gamma+log n)+frac{n}{loglog n}}$$
$H(n)$ tends to $e^gamma n$, which is the same as $gamma + log n$. So, you can reword it as this:
$$lim_{nrightarrowinfty}frac{e^{gamma + log n}log(gamma + log n)}{(gamma + log n)log(gamma+log n)+frac{n}{loglog n}}$$
So how do you continue to prove that this limit is $1$?
calculus limits
asked 10 hours ago
Quote DaveQuote Dave
12111
$endgroup$
1
$begingroup$
It seems that you are assuming that $$lim_{xtoinfty}frac{h(x)}{f(x)g(x)}=0$$ for all functions $f$, $g$, $h$.
$endgroup$
– Peter Foreman
10 hours ago
$begingroup$
Ok, I have a specific limit in mid, let me pull it up.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
@QuoteDave Perhaps you should ask a new question; with your edit, your question becomes quite different from what you originally asked.
$endgroup$
– Sambo
10 hours ago
$begingroup$
$mathrm{e}^gamma n neq gamma + ln n$. In particular, try $gamma = 0$, $n = 1$. What is true is that $mathrm{e}^gamma n = mathrm{e}^{gamma + ln n}$ for $n > 0$.
$endgroup$
– Eric Towers
10 hours ago
|
show 7 more comments
$begingroup$
I have found that for distinct functions (not constants) $f(x),g(x),h(x)$ that this is true:
$$lim_{x to infty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=1$$ I want to know: why is that, ie is there a rigorous proof for why that is? Maybe a paper to turn to or is it really simple?
EDIT: I realized I probably should have put my actual problem first. I hope this gives more context. For this limit:
$$lim_{nrightarrowinfty}frac{e^{H(n)}log H(n)}{e^gamma nlog(gamma+log n)+frac{n}{loglog n}}$$
$H(n)$ tends to $e^gamma n$, which is the same as $gamma + log n$. So, you can reword it as this:
$$lim_{nrightarrowinfty}frac{e^{gamma + log n}log(gamma + log n)}{(gamma + log n)log(gamma+log n)+frac{n}{loglog n}}$$
So how do you continue to prove that this limit is $1$?
calculus limits
asked 10 hours ago
Quote DaveQuote Dave
12111
$endgroup$
I have found that for distinct functions (not constants) $f(x),g(x),h(x)$ that this is true:
$$lim_{x to infty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=1$$ I want to know: why is that, ie is there a rigorous proof for why that is? Maybe a paper to turn to or is it really simple?
EDIT: I realized I probably should have put my actual problem first. I hope this gives more context. For this limit:
$$lim_{nrightarrowinfty}frac{e^{H(n)}log H(n)}{e^gamma nlog(gamma+log n)+frac{n}{loglog n}}$$
$H(n)$ tends to $e^gamma n$, which is the same as $gamma + log n$. So, you can reword it as this:
$$lim_{nrightarrowinfty}frac{e^{gamma + log n}log(gamma + log n)}{(gamma + log n)log(gamma+log n)+frac{n}{loglog n}}$$
So how do you continue to prove that this limit is $1$?
calculus limits
calculus limits
asked 10 hours ago
Quote DaveQuote Dave
12111
asked 10 hours ago
Quote DaveQuote Dave
12111
edited 10 hours ago
Quote Dave
asked 10 hours ago
Quote DaveQuote Dave
12111
asked 10 hours ago
Quote DaveQuote Dave
12111
asked 10 hours ago
Quote DaveQuote Dave
12111
12111
1
$begingroup$
It seems that you are assuming that $$lim_{xtoinfty}frac{h(x)}{f(x)g(x)}=0$$ for all functions $f$, $g$, $h$.
$endgroup$
– Peter Foreman
10 hours ago
$begingroup$
Ok, I have a specific limit in mid, let me pull it up.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
@QuoteDave Perhaps you should ask a new question; with your edit, your question becomes quite different from what you originally asked.
$endgroup$
– Sambo
10 hours ago
$begingroup$
$mathrm{e}^gamma n neq gamma + ln n$. In particular, try $gamma = 0$, $n = 1$. What is true is that $mathrm{e}^gamma n = mathrm{e}^{gamma + ln n}$ for $n > 0$.
$endgroup$
– Eric Towers
10 hours ago
|
show 7 more comments
1
$begingroup$
It seems that you are assuming that $$lim_{xtoinfty}frac{h(x)}{f(x)g(x)}=0$$ for all functions $f$, $g$, $h$.
$endgroup$
– Peter Foreman
10 hours ago
$begingroup$
Ok, I have a specific limit in mid, let me pull it up.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
@QuoteDave Perhaps you should ask a new question; with your edit, your question becomes quite different from what you originally asked.
$endgroup$
– Sambo
10 hours ago
$begingroup$
$mathrm{e}^gamma n neq gamma + ln n$. In particular, try $gamma = 0$, $n = 1$. What is true is that $mathrm{e}^gamma n = mathrm{e}^{gamma + ln n}$ for $n > 0$.
$endgroup$
– Eric Towers
10 hours ago
1
1
$begingroup$
It seems that you are assuming that $$lim_{xtoinfty}frac{h(x)}{f(x)g(x)}=0$$ for all functions $f$, $g$, $h$.
$endgroup$
– Peter Foreman
10 hours ago
$begingroup$
It seems that you are assuming that $$lim_{xtoinfty}frac{h(x)}{f(x)g(x)}=0$$ for all functions $f$, $g$, $h$.
$endgroup$
– Peter Foreman
10 hours ago
$begingroup$
Ok, I have a specific limit in mid, let me pull it up.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I have a specific limit in mid, let me pull it up.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
@QuoteDave Perhaps you should ask a new question; with your edit, your question becomes quite different from what you originally asked.
$endgroup$
– Sambo
10 hours ago
$begingroup$
@QuoteDave Perhaps you should ask a new question; with your edit, your question becomes quite different from what you originally asked.
$endgroup$
– Sambo
10 hours ago
$begingroup$
$mathrm{e}^gamma n neq gamma + ln n$. In particular, try $gamma = 0$, $n = 1$. What is true is that $mathrm{e}^gamma n = mathrm{e}^{gamma + ln n}$ for $n > 0$.
$endgroup$
– Eric Towers
10 hours ago
$begingroup$
$mathrm{e}^gamma n neq gamma + ln n$. In particular, try $gamma = 0$, $n = 1$. What is true is that $mathrm{e}^gamma n = mathrm{e}^{gamma + ln n}$ for $n > 0$.
$endgroup$
– Eric Towers
10 hours ago
|
show 7 more comments
4 Answers
4
active
oldest
votes
$begingroup$
There are many continuous nonconstant functions whose limit is zero. For instance, using three of these, let $f(x) = dfrac{1}{x}$, $g(x) = mathrm{e}^{-x}$, and $h(x) = 1+ dfrac{sin x}{x}$ so that
$$ lim_{x rightarrow infty} frac{f(x)g(x)}{f(x)g(x) + h(x)} = frac{0 cdot 0}{0 cdot 0 + 1} = 0 text{.} $$
Anticipating where your edit is going. Use the manipulation
$$ frac{f}{f+frac{n}{ln ln n}} = frac{f}{f+frac{n}{ln ln n}} cdot frac{, frac{1}{f} ,}{frac{1}{f}} = frac{1}{1 + frac{n}{f ln ln n}} text{.} $$
Since your $f$ grows slightly faster than linearly in $n$, you get $dfrac{1}{1+0}$ as your limit.
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
$endgroup$
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
If $f(x)=x$, $g(x)=x^2$, and $h(x)=x^4$, then$$lim_{xtoinfty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=0.$$So, the statement is false.
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
5
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
8
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
add a comment |
$begingroup$
This statement is false as pointed out by Jose.
For example, if I pick $h(x) = f(x)g(x) $, then
$$ lim_{x to infty} frac{f(x) g(x)}{ f(x) g(x) + f(x) g(x)} = frac{1}{2} $$
no matter if you have constant or not, it will always be $frac{1}{2}$
answered 10 hours ago
user209663user209663
1,004411
$endgroup$
add a comment |
$begingroup$
This is not true in general. For example, if $f$, $g$, and $h$ are all the constant function that maps to $1$, the limit is $1/2$. If you want distinct functions, take $f=1, g=2, h=3$ uniformly to see that the limit is $2/5neq 1$.
On the other hand, if $h$ is $o(fg)$ and $fg$ is nonzero, then $$lim_{xtoinfty} frac{f(x) g(x)}{f(x)g(x) + h(x)} = lim_{xtoinfty} frac{1}{1 + h(x) / (f(x)g(x))} = 1. $$
In that situation $fg$ is nonzero, the above argument says that the limit is one if and only if $h = o(fg)$.
answered 10 hours ago
cdipaolocdipaolo
705313
$endgroup$
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are many continuous nonconstant functions whose limit is zero. For instance, using three of these, let $f(x) = dfrac{1}{x}$, $g(x) = mathrm{e}^{-x}$, and $h(x) = 1+ dfrac{sin x}{x}$ so that
$$ lim_{x rightarrow infty} frac{f(x)g(x)}{f(x)g(x) + h(x)} = frac{0 cdot 0}{0 cdot 0 + 1} = 0 text{.} $$
Anticipating where your edit is going. Use the manipulation
$$ frac{f}{f+frac{n}{ln ln n}} = frac{f}{f+frac{n}{ln ln n}} cdot frac{, frac{1}{f} ,}{frac{1}{f}} = frac{1}{1 + frac{n}{f ln ln n}} text{.} $$
Since your $f$ grows slightly faster than linearly in $n$, you get $dfrac{1}{1+0}$ as your limit.
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
$endgroup$
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
There are many continuous nonconstant functions whose limit is zero. For instance, using three of these, let $f(x) = dfrac{1}{x}$, $g(x) = mathrm{e}^{-x}$, and $h(x) = 1+ dfrac{sin x}{x}$ so that
$$ lim_{x rightarrow infty} frac{f(x)g(x)}{f(x)g(x) + h(x)} = frac{0 cdot 0}{0 cdot 0 + 1} = 0 text{.} $$
Anticipating where your edit is going. Use the manipulation
$$ frac{f}{f+frac{n}{ln ln n}} = frac{f}{f+frac{n}{ln ln n}} cdot frac{, frac{1}{f} ,}{frac{1}{f}} = frac{1}{1 + frac{n}{f ln ln n}} text{.} $$
Since your $f$ grows slightly faster than linearly in $n$, you get $dfrac{1}{1+0}$ as your limit.
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
$endgroup$
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
There are many continuous nonconstant functions whose limit is zero. For instance, using three of these, let $f(x) = dfrac{1}{x}$, $g(x) = mathrm{e}^{-x}$, and $h(x) = 1+ dfrac{sin x}{x}$ so that
$$ lim_{x rightarrow infty} frac{f(x)g(x)}{f(x)g(x) + h(x)} = frac{0 cdot 0}{0 cdot 0 + 1} = 0 text{.} $$
Anticipating where your edit is going. Use the manipulation
$$ frac{f}{f+frac{n}{ln ln n}} = frac{f}{f+frac{n}{ln ln n}} cdot frac{, frac{1}{f} ,}{frac{1}{f}} = frac{1}{1 + frac{n}{f ln ln n}} text{.} $$
Since your $f$ grows slightly faster than linearly in $n$, you get $dfrac{1}{1+0}$ as your limit.
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
$endgroup$
There are many continuous nonconstant functions whose limit is zero. For instance, using three of these, let $f(x) = dfrac{1}{x}$, $g(x) = mathrm{e}^{-x}$, and $h(x) = 1+ dfrac{sin x}{x}$ so that
$$ lim_{x rightarrow infty} frac{f(x)g(x)}{f(x)g(x) + h(x)} = frac{0 cdot 0}{0 cdot 0 + 1} = 0 text{.} $$
Anticipating where your edit is going. Use the manipulation
$$ frac{f}{f+frac{n}{ln ln n}} = frac{f}{f+frac{n}{ln ln n}} cdot frac{, frac{1}{f} ,}{frac{1}{f}} = frac{1}{1 + frac{n}{f ln ln n}} text{.} $$
Since your $f$ grows slightly faster than linearly in $n$, you get $dfrac{1}{1+0}$ as your limit.
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
edited 10 hours ago
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
answered 10 hours ago
Eric TowersEric Towers
34.8k22371
34.8k22371
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
If $f(x)=x$, $g(x)=x^2$, and $h(x)=x^4$, then$$lim_{xtoinfty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=0.$$So, the statement is false.
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
5
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
8
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
add a comment |
$begingroup$
If $f(x)=x$, $g(x)=x^2$, and $h(x)=x^4$, then$$lim_{xtoinfty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=0.$$So, the statement is false.
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
5
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
8
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
add a comment |
$begingroup$
If $f(x)=x$, $g(x)=x^2$, and $h(x)=x^4$, then$$lim_{xtoinfty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=0.$$So, the statement is false.
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
If $f(x)=x$, $g(x)=x^2$, and $h(x)=x^4$, then$$lim_{xtoinfty}frac{f(x)g(x)}{f(x)g(x)+h(x)}=0.$$So, the statement is false.
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
edited 10 hours ago
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
193k24148266
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
5
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
8
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
add a comment |
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
5
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
8
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
I changed it to include non-constants.
$endgroup$
– Quote Dave
10 hours ago
5
5
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
And I've edited my answer.
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
8
8
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
$begingroup$
How can someone answer your question if you keep changing it?
$endgroup$
– José Carlos Santos
10 hours ago
add a comment |
$begingroup$
This statement is false as pointed out by Jose.
For example, if I pick $h(x) = f(x)g(x) $, then
$$ lim_{x to infty} frac{f(x) g(x)}{ f(x) g(x) + f(x) g(x)} = frac{1}{2} $$
no matter if you have constant or not, it will always be $frac{1}{2}$
answered 10 hours ago
user209663user209663
1,004411
$endgroup$
add a comment |
$begingroup$
This statement is false as pointed out by Jose.
For example, if I pick $h(x) = f(x)g(x) $, then
$$ lim_{x to infty} frac{f(x) g(x)}{ f(x) g(x) + f(x) g(x)} = frac{1}{2} $$
no matter if you have constant or not, it will always be $frac{1}{2}$
answered 10 hours ago
user209663user209663
1,004411
$endgroup$
add a comment |
$begingroup$
This statement is false as pointed out by Jose.
For example, if I pick $h(x) = f(x)g(x) $, then
$$ lim_{x to infty} frac{f(x) g(x)}{ f(x) g(x) + f(x) g(x)} = frac{1}{2} $$
no matter if you have constant or not, it will always be $frac{1}{2}$
answered 10 hours ago
user209663user209663
1,004411
$endgroup$
This statement is false as pointed out by Jose.
For example, if I pick $h(x) = f(x)g(x) $, then
$$ lim_{x to infty} frac{f(x) g(x)}{ f(x) g(x) + f(x) g(x)} = frac{1}{2} $$
no matter if you have constant or not, it will always be $frac{1}{2}$
answered 10 hours ago
user209663user209663
1,004411
answered 10 hours ago
user209663user209663
1,004411
answered 10 hours ago
user209663user209663
1,004411
answered 10 hours ago
user209663user209663
1,004411
1,004411
add a comment |
add a comment |
$begingroup$
This is not true in general. For example, if $f$, $g$, and $h$ are all the constant function that maps to $1$, the limit is $1/2$. If you want distinct functions, take $f=1, g=2, h=3$ uniformly to see that the limit is $2/5neq 1$.
On the other hand, if $h$ is $o(fg)$ and $fg$ is nonzero, then $$lim_{xtoinfty} frac{f(x) g(x)}{f(x)g(x) + h(x)} = lim_{xtoinfty} frac{1}{1 + h(x) / (f(x)g(x))} = 1. $$
In that situation $fg$ is nonzero, the above argument says that the limit is one if and only if $h = o(fg)$.
answered 10 hours ago
cdipaolocdipaolo
705313
$endgroup$
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
This is not true in general. For example, if $f$, $g$, and $h$ are all the constant function that maps to $1$, the limit is $1/2$. If you want distinct functions, take $f=1, g=2, h=3$ uniformly to see that the limit is $2/5neq 1$.
On the other hand, if $h$ is $o(fg)$ and $fg$ is nonzero, then $$lim_{xtoinfty} frac{f(x) g(x)}{f(x)g(x) + h(x)} = lim_{xtoinfty} frac{1}{1 + h(x) / (f(x)g(x))} = 1. $$
In that situation $fg$ is nonzero, the above argument says that the limit is one if and only if $h = o(fg)$.
answered 10 hours ago
cdipaolocdipaolo
705313
$endgroup$
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
This is not true in general. For example, if $f$, $g$, and $h$ are all the constant function that maps to $1$, the limit is $1/2$. If you want distinct functions, take $f=1, g=2, h=3$ uniformly to see that the limit is $2/5neq 1$.
On the other hand, if $h$ is $o(fg)$ and $fg$ is nonzero, then $$lim_{xtoinfty} frac{f(x) g(x)}{f(x)g(x) + h(x)} = lim_{xtoinfty} frac{1}{1 + h(x) / (f(x)g(x))} = 1. $$
In that situation $fg$ is nonzero, the above argument says that the limit is one if and only if $h = o(fg)$.
answered 10 hours ago
cdipaolocdipaolo
705313
$endgroup$
This is not true in general. For example, if $f$, $g$, and $h$ are all the constant function that maps to $1$, the limit is $1/2$. If you want distinct functions, take $f=1, g=2, h=3$ uniformly to see that the limit is $2/5neq 1$.
On the other hand, if $h$ is $o(fg)$ and $fg$ is nonzero, then $$lim_{xtoinfty} frac{f(x) g(x)}{f(x)g(x) + h(x)} = lim_{xtoinfty} frac{1}{1 + h(x) / (f(x)g(x))} = 1. $$
In that situation $fg$ is nonzero, the above argument says that the limit is one if and only if $h = o(fg)$.
answered 10 hours ago
cdipaolocdipaolo
705313
answered 10 hours ago
cdipaolocdipaolo
705313
answered 10 hours ago
cdipaolocdipaolo
705313
answered 10 hours ago
cdipaolocdipaolo
705313
705313
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
add a comment |
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It seems that you are assuming that $$lim_{xtoinfty}frac{h(x)}{f(x)g(x)}=0$$ for all functions $f$, $g$, $h$.
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– Peter Foreman
10 hours ago
$begingroup$
Ok, I have a specific limit in mid, let me pull it up.
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– Quote Dave
10 hours ago
$begingroup$
Ok, I edited my question.
$endgroup$
– Quote Dave
10 hours ago
$begingroup$
@QuoteDave Perhaps you should ask a new question; with your edit, your question becomes quite different from what you originally asked.
$endgroup$
– Sambo
10 hours ago
$begingroup$
$mathrm{e}^gamma n neq gamma + ln n$. In particular, try $gamma = 0$, $n = 1$. What is true is that $mathrm{e}^gamma n = mathrm{e}^{gamma + ln n}$ for $n > 0$.
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– Eric Towers
10 hours ago