Mixing
Calculate: $limlimits_{xto0^-} frac1{ln(1-x)}+frac1x$ without LHR/Expansions [duplicate]
$begingroup$
This question already has an answer here:
Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$
4 answers
How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?
I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-frac{x^2}2-x-1$
We get:
$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$
Thus:
$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$
By substituting: $t=ln(1-x)$ in the original limit we get:
$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$
limits logarithms exponential-function
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
$endgroup$
marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$
4 answers
How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?
I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-frac{x^2}2-x-1$
We get:
$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$
Thus:
$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$
By substituting: $t=ln(1-x)$ in the original limit we get:
$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$
limits logarithms exponential-function
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
$endgroup$
marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52
$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26
add a comment |
$begingroup$
This question already has an answer here:
Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$
4 answers
How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?
I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-frac{x^2}2-x-1$
We get:
$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$
Thus:
$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$
By substituting: $t=ln(1-x)$ in the original limit we get:
$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$
limits logarithms exponential-function
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
$endgroup$
This question already has an answer here:
Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$
4 answers
How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?
I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-frac{x^2}2-x-1$
We get:
$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$
Thus:
$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$
By substituting: $t=ln(1-x)$ in the original limit we get:
$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$
This question already has an answer here:
Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$
4 answers
limits logarithms exponential-function
limits logarithms exponential-function
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
edited Jan 5 at 19:12
Bilal HIMITE
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
asked Jan 5 at 15:20
Bilal HIMITEBilal HIMITE
806
806
marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52
$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26
add a comment |
$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52
$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26
$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52
$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52
$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26
$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ This gives $L=frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$
answered Jan 5 at 15:51
SongSong
9,644627
$endgroup$
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
add a comment |
$begingroup$
First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$
Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$
answered Jan 5 at 16:02
BernardBernard
119k740113
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ This gives $L=frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$
answered Jan 5 at 15:51
SongSong
9,644627
$endgroup$
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
add a comment |
$begingroup$
Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ This gives $L=frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$
answered Jan 5 at 15:51
SongSong
9,644627
$endgroup$
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
add a comment |
$begingroup$
Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ This gives $L=frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$
answered Jan 5 at 15:51
SongSong
9,644627
$endgroup$
Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ This gives $L=frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$
answered Jan 5 at 15:51
SongSong
9,644627
edited Jan 5 at 16:11
answered Jan 5 at 15:51
SongSong
9,644627
answered Jan 5 at 15:51
SongSong
9,644627
answered Jan 5 at 15:51
SongSong
9,644627
9,644627
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
add a comment |
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
The last steps are brilliant (+1)
$endgroup$
– TheSimpliFire
Jan 5 at 15:52
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
$endgroup$
– Bilal HIMITE
Jan 5 at 18:33
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
$begingroup$
Well, that is why I added the 'justication' part at the end. Please chech that.
$endgroup$
– Song
Jan 5 at 18:35
add a comment |
$begingroup$
First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$
Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$
answered Jan 5 at 16:02
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$
Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$
answered Jan 5 at 16:02
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$
Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$
answered Jan 5 at 16:02
BernardBernard
119k740113
$endgroup$
First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$
Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$
answered Jan 5 at 16:02
BernardBernard
119k740113
answered Jan 5 at 16:02
BernardBernard
119k740113
answered Jan 5 at 16:02
BernardBernard
119k740113
answered Jan 5 at 16:02
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52
$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26