Limlit of $frac{x-xln(1+x)-ln(1+x)}{x^3+x^2}$ at $0$ without l'Hôpital or Taylor series
I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)
logarithms limits-without-lhopital indeterminate-forms
edited yesterday
mrtaurho
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asked yesterday
Heryon
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I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)
logarithms limits-without-lhopital indeterminate-forms
edited yesterday
mrtaurho
4,01621133
asked yesterday
Heryon
194
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Heryon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)
logarithms limits-without-lhopital indeterminate-forms
edited yesterday
mrtaurho
4,01621133
asked yesterday
Heryon
194
New contributor
Heryon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)
logarithms limits-without-lhopital indeterminate-forms
logarithms limits-without-lhopital indeterminate-forms
edited yesterday
mrtaurho
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asked yesterday
Heryon
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edited yesterday
mrtaurho
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asked yesterday
Heryon
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edited yesterday
mrtaurho
4,01621133
edited yesterday
mrtaurho
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asked yesterday
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4 Answers
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First, you can decompose the first term in your derivative as:
$frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$frac{ln(1+x)}{x^2}$
$ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$
$ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $
Putting it together, your derivative is:
$frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$
- The $1/x$ terms cancel out.
$lim_{x rightarrow 0} frac{1}{1+x} = 1$.
$lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.
So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.
answered yesterday
Aditya Dua
86418
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
add a comment |
Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
$$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
$$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
$$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
So,
$$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$
answered yesterday
Peter Foreman
2056
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$frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$
At which point I am inclined to use a series
$frac {d}{dx} (1 - frac x2 + cdots)$
as $x$ goes to $0$ equals $-frac 12$
Without using a series, I have.
$frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$
is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$
answered yesterday
Doug M
44.2k31854
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
add a comment |
Because $frac{x}{1+x}<ln(1+x)<x$
So when $ x rightarrow 0+, f(x) rightarrow 1$
When $xrightarrow0-$ is similar
answered yesterday
李子镔
314
add a comment |
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4 Answers
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active
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4 Answers
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First, you can decompose the first term in your derivative as:
$frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$frac{ln(1+x)}{x^2}$
$ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$
$ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $
Putting it together, your derivative is:
$frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$
- The $1/x$ terms cancel out.
$lim_{x rightarrow 0} frac{1}{1+x} = 1$.
$lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.
So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.
answered yesterday
Aditya Dua
86418
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
add a comment |
First, you can decompose the first term in your derivative as:
$frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$frac{ln(1+x)}{x^2}$
$ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$
$ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $
Putting it together, your derivative is:
$frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$
- The $1/x$ terms cancel out.
$lim_{x rightarrow 0} frac{1}{1+x} = 1$.
$lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.
So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.
answered yesterday
Aditya Dua
86418
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
add a comment |
First, you can decompose the first term in your derivative as:
$frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$frac{ln(1+x)}{x^2}$
$ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$
$ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $
Putting it together, your derivative is:
$frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$
- The $1/x$ terms cancel out.
$lim_{x rightarrow 0} frac{1}{1+x} = 1$.
$lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.
So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.
answered yesterday
Aditya Dua
86418
First, you can decompose the first term in your derivative as:
$frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$frac{ln(1+x)}{x^2}$
$ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$
$ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $
Putting it together, your derivative is:
$frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$
- The $1/x$ terms cancel out.
$lim_{x rightarrow 0} frac{1}{1+x} = 1$.
$lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.
So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.
answered yesterday
Aditya Dua
86418
answered yesterday
Aditya Dua
86418
answered yesterday
Aditya Dua
86418
answered yesterday
Aditya Dua
86418
86418
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
add a comment |
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
Clever, thanks. Any way to do this without Taylor series? (Just to know)
– Heryon
yesterday
add a comment |
Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
$$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
$$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
$$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
So,
$$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$
answered yesterday
Peter Foreman
2056
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Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
$$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
$$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
$$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
So,
$$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$
answered yesterday
Peter Foreman
2056
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Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
$$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
$$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
$$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
So,
$$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$
answered yesterday
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
$$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
$$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
$$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
So,
$$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$
answered yesterday
Peter Foreman
2056
New contributor
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answered yesterday
Peter Foreman
2056
New contributor
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answered yesterday
Peter Foreman
2056
answered yesterday
Peter Foreman
2056
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
$frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$
At which point I am inclined to use a series
$frac {d}{dx} (1 - frac x2 + cdots)$
as $x$ goes to $0$ equals $-frac 12$
Without using a series, I have.
$frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$
is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$
answered yesterday
Doug M
44.2k31854
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
add a comment |
$frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$
At which point I am inclined to use a series
$frac {d}{dx} (1 - frac x2 + cdots)$
as $x$ goes to $0$ equals $-frac 12$
Without using a series, I have.
$frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$
is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$
answered yesterday
Doug M
44.2k31854
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
add a comment |
$frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$
At which point I am inclined to use a series
$frac {d}{dx} (1 - frac x2 + cdots)$
as $x$ goes to $0$ equals $-frac 12$
Without using a series, I have.
$frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$
is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$
answered yesterday
Doug M
44.2k31854
$frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$
At which point I am inclined to use a series
$frac {d}{dx} (1 - frac x2 + cdots)$
as $x$ goes to $0$ equals $-frac 12$
Without using a series, I have.
$frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$
is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$
answered yesterday
Doug M
44.2k31854
edited yesterday
answered yesterday
Doug M
44.2k31854
answered yesterday
Doug M
44.2k31854
answered yesterday
Doug M
44.2k31854
44.2k31854
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
add a comment |
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
Yes I can't use that too, edited the post, but thanks!
– Heryon
yesterday
add a comment |
Because $frac{x}{1+x}<ln(1+x)<x$
So when $ x rightarrow 0+, f(x) rightarrow 1$
When $xrightarrow0-$ is similar
answered yesterday
李子镔
314
add a comment |
Because $frac{x}{1+x}<ln(1+x)<x$
So when $ x rightarrow 0+, f(x) rightarrow 1$
When $xrightarrow0-$ is similar
answered yesterday
李子镔
314
add a comment |
Because $frac{x}{1+x}<ln(1+x)<x$
So when $ x rightarrow 0+, f(x) rightarrow 1$
When $xrightarrow0-$ is similar
answered yesterday
李子镔
314
Because $frac{x}{1+x}<ln(1+x)<x$
So when $ x rightarrow 0+, f(x) rightarrow 1$
When $xrightarrow0-$ is similar
answered yesterday
李子镔
314
answered yesterday
李子镔
314
answered yesterday
李子镔
314
answered yesterday
李子镔
314
314
add a comment |
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
