How to develop a solution in limit without L'Hopital?
Find $displaystylelim_{xto1}left(dfrac{1}{1-x}-dfrac{1}{ln x}right).$
I don't know how to remove $infty-infty$ uncertainty in the question. Please explain it and how many different solution can we do in the question?
calculus limits
asked Dec 9 '18 at 6:02
Atakan
184
|
show 12 more comments
Find $displaystylelim_{xto1}left(dfrac{1}{1-x}-dfrac{1}{ln x}right).$
I don't know how to remove $infty-infty$ uncertainty in the question. Please explain it and how many different solution can we do in the question?
calculus limits
asked Dec 9 '18 at 6:02
Atakan
184
@copper.hat I put a picture link into sentence.You can look at it
– Atakan
Dec 9 '18 at 6:12
Are you familiar with L'Hôpital's rule? (The rule can be repeated as necessary.)
– copper.hat
Dec 9 '18 at 6:25
1
@Atakan: Limit does not exist check by $xto 1^-$ and $xto1^+$.
– Yadati Kiran
Dec 9 '18 at 6:28
1
@AniruddhVenkatesan: Yeah. as $xto1^-$ we have $+infty$ and $xto1^+$ we have $-infty$
– Yadati Kiran
Dec 9 '18 at 6:31
1
@Atakan: The two sided limit does not exist.
– Yadati Kiran
Dec 9 '18 at 6:34
|
show 12 more comments
Find $displaystylelim_{xto1}left(dfrac{1}{1-x}-dfrac{1}{ln x}right).$
I don't know how to remove $infty-infty$ uncertainty in the question. Please explain it and how many different solution can we do in the question?
calculus limits
asked Dec 9 '18 at 6:02
Atakan
184
Find $displaystylelim_{xto1}left(dfrac{1}{1-x}-dfrac{1}{ln x}right).$
I don't know how to remove $infty-infty$ uncertainty in the question. Please explain it and how many different solution can we do in the question?
calculus limits
calculus limits
asked Dec 9 '18 at 6:02
Atakan
184
asked Dec 9 '18 at 6:02
Atakan
184
edited Dec 9 '18 at 7:20
asked Dec 9 '18 at 6:02
Atakan
184
asked Dec 9 '18 at 6:02
Atakan
184
asked Dec 9 '18 at 6:02
Atakan
184
184
@copper.hat I put a picture link into sentence.You can look at it
– Atakan
Dec 9 '18 at 6:12
Are you familiar with L'Hôpital's rule? (The rule can be repeated as necessary.)
– copper.hat
Dec 9 '18 at 6:25
1
@Atakan: Limit does not exist check by $xto 1^-$ and $xto1^+$.
– Yadati Kiran
Dec 9 '18 at 6:28
1
@AniruddhVenkatesan: Yeah. as $xto1^-$ we have $+infty$ and $xto1^+$ we have $-infty$
– Yadati Kiran
Dec 9 '18 at 6:31
1
@Atakan: The two sided limit does not exist.
– Yadati Kiran
Dec 9 '18 at 6:34
|
show 12 more comments
@copper.hat I put a picture link into sentence.You can look at it
– Atakan
Dec 9 '18 at 6:12
Are you familiar with L'Hôpital's rule? (The rule can be repeated as necessary.)
– copper.hat
Dec 9 '18 at 6:25
1
@Atakan: Limit does not exist check by $xto 1^-$ and $xto1^+$.
– Yadati Kiran
Dec 9 '18 at 6:28
1
@AniruddhVenkatesan: Yeah. as $xto1^-$ we have $+infty$ and $xto1^+$ we have $-infty$
– Yadati Kiran
Dec 9 '18 at 6:31
1
@Atakan: The two sided limit does not exist.
– Yadati Kiran
Dec 9 '18 at 6:34
@copper.hat I put a picture link into sentence.You can look at it
– Atakan
Dec 9 '18 at 6:12
@copper.hat I put a picture link into sentence.You can look at it
– Atakan
Dec 9 '18 at 6:12
Are you familiar with L'Hôpital's rule? (The rule can be repeated as necessary.)
– copper.hat
Dec 9 '18 at 6:25
Are you familiar with L'Hôpital's rule? (The rule can be repeated as necessary.)
– copper.hat
Dec 9 '18 at 6:25
1
1
@Atakan: Limit does not exist check by $xto 1^-$ and $xto1^+$.
– Yadati Kiran
Dec 9 '18 at 6:28
@Atakan: Limit does not exist check by $xto 1^-$ and $xto1^+$.
– Yadati Kiran
Dec 9 '18 at 6:28
1
1
@AniruddhVenkatesan: Yeah. as $xto1^-$ we have $+infty$ and $xto1^+$ we have $-infty$
– Yadati Kiran
Dec 9 '18 at 6:31
@AniruddhVenkatesan: Yeah. as $xto1^-$ we have $+infty$ and $xto1^+$ we have $-infty$
– Yadati Kiran
Dec 9 '18 at 6:31
1
1
@Atakan: The two sided limit does not exist.
– Yadati Kiran
Dec 9 '18 at 6:34
@Atakan: The two sided limit does not exist.
– Yadati Kiran
Dec 9 '18 at 6:34
|
show 12 more comments
5 Answers
5
active
oldest
votes
You could try to use Taylor Series expansions.
$displaystylelim_{mto0}frac1m-frac1{ln(1-m)}, m=1-x$
$=displaystylelim_{mto0}frac1m+frac1{m+frac{m^2}2+frac{m^3}{3}...}=displaystylelim_{mto0}frac1m(1+frac1{1+frac m2+frac{m^2}3...})$
which diverges to $-infty$ when $mto0^-implies xto1^+$, and $+infty$ when $mto0^+implies xto1^-$.
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
1
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
1
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
|
show 4 more comments
Write ${1 over 1-x} - {1 over log x} = {log x +x-1over (1-x) log x}$.
The Taylor expansions are
$log x +x-1 = 2(x-1)+cdots$ and $(1-x) log x = -(x-1)^2+ cdots $.
Hence we expect the behaviour for $x$ close to $1$ to be $approx -{2 over x-1}$.
answered Dec 9 '18 at 6:56
copper.hat
126k559159
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
add a comment |
I want to rewrite the limit slightly:
$$lim_{xto1} left(frac{1}{1-x} - frac{1}{ln(x)}right) = -lim_{xto1} left(frac{1}{ln(x)} + frac{1}{x-1} right). $$
Note that now our two summands have the same sign when $x<1$ and when $x>1$. The limit from the left goes to $(-infty)+ (-infty) = (-infty)$ and the limit from the right goes to $infty + infty = infty$. Thus, the limit does not exist.
No indefinite forms needed!
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
add a comment |
I guess the first term should have been $frac{1}{x-1}$ rather than $frac{1}{1-x}$. In the former case the substitution $x=e^t$ reduces the problem to the evaluation of
$$ lim_{tto 0}left(frac{1}{e^t-1}-frac{1}{t}right)=-lim_{t to 0}frac{e^t-1-t}{t(e^t-1)}=-lim_{tto 0}frac{frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=color{red}{-frac{1}{2}}. $$
The Maclaurin series of $e^t$ is either a trivial consequence of the series definition of $e^t$ or a simple consequence of the dominated convergence theorem / integration by parts:
$$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
add a comment |
We have that by $y=x-1 to 0$
$$lim_{xto1} left(dfrac{1}{1-x}-dfrac{1}{ln x}right) =lim_{yto0} left(-dfrac{1}{y}-dfrac{1}{ln (1+y)}right)$$
and by standard limits since $frac{ln(1+y)}yto 1$
$$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$
answered Dec 9 '18 at 8:12
gimusi
1
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could try to use Taylor Series expansions.
$displaystylelim_{mto0}frac1m-frac1{ln(1-m)}, m=1-x$
$=displaystylelim_{mto0}frac1m+frac1{m+frac{m^2}2+frac{m^3}{3}...}=displaystylelim_{mto0}frac1m(1+frac1{1+frac m2+frac{m^2}3...})$
which diverges to $-infty$ when $mto0^-implies xto1^+$, and $+infty$ when $mto0^+implies xto1^-$.
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
1
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
1
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
|
show 4 more comments
You could try to use Taylor Series expansions.
$displaystylelim_{mto0}frac1m-frac1{ln(1-m)}, m=1-x$
$=displaystylelim_{mto0}frac1m+frac1{m+frac{m^2}2+frac{m^3}{3}...}=displaystylelim_{mto0}frac1m(1+frac1{1+frac m2+frac{m^2}3...})$
which diverges to $-infty$ when $mto0^-implies xto1^+$, and $+infty$ when $mto0^+implies xto1^-$.
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
1
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
1
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
|
show 4 more comments
You could try to use Taylor Series expansions.
$displaystylelim_{mto0}frac1m-frac1{ln(1-m)}, m=1-x$
$=displaystylelim_{mto0}frac1m+frac1{m+frac{m^2}2+frac{m^3}{3}...}=displaystylelim_{mto0}frac1m(1+frac1{1+frac m2+frac{m^2}3...})$
which diverges to $-infty$ when $mto0^-implies xto1^+$, and $+infty$ when $mto0^+implies xto1^-$.
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
You could try to use Taylor Series expansions.
$displaystylelim_{mto0}frac1m-frac1{ln(1-m)}, m=1-x$
$=displaystylelim_{mto0}frac1m+frac1{m+frac{m^2}2+frac{m^3}{3}...}=displaystylelim_{mto0}frac1m(1+frac1{1+frac m2+frac{m^2}3...})$
which diverges to $-infty$ when $mto0^-implies xto1^+$, and $+infty$ when $mto0^+implies xto1^-$.
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
edited Dec 9 '18 at 12:26
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
answered Dec 9 '18 at 6:37
Shubham Johri
3,961717
3,961717
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
1
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
1
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
|
show 4 more comments
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
1
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
1
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
Ohh,I liked your solution and I'm supposing the book is wrong about shown answer.Because your solution is looking true
– Atakan
Dec 9 '18 at 6:43
1
1
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
The graph of $frac1{1-x}-frac1{ln x}$ supports my observation.
– Shubham Johri
Dec 9 '18 at 6:48
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
how did you prove it on the graph?Can you show it on the graph?
– Atakan
Dec 9 '18 at 6:51
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
thank you for showing it :)
– Atakan
Dec 9 '18 at 6:52
1
1
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
You're welcome.
– Shubham Johri
Dec 9 '18 at 6:54
|
show 4 more comments
Write ${1 over 1-x} - {1 over log x} = {log x +x-1over (1-x) log x}$.
The Taylor expansions are
$log x +x-1 = 2(x-1)+cdots$ and $(1-x) log x = -(x-1)^2+ cdots $.
Hence we expect the behaviour for $x$ close to $1$ to be $approx -{2 over x-1}$.
answered Dec 9 '18 at 6:56
copper.hat
126k559159
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
add a comment |
Write ${1 over 1-x} - {1 over log x} = {log x +x-1over (1-x) log x}$.
The Taylor expansions are
$log x +x-1 = 2(x-1)+cdots$ and $(1-x) log x = -(x-1)^2+ cdots $.
Hence we expect the behaviour for $x$ close to $1$ to be $approx -{2 over x-1}$.
answered Dec 9 '18 at 6:56
copper.hat
126k559159
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
add a comment |
Write ${1 over 1-x} - {1 over log x} = {log x +x-1over (1-x) log x}$.
The Taylor expansions are
$log x +x-1 = 2(x-1)+cdots$ and $(1-x) log x = -(x-1)^2+ cdots $.
Hence we expect the behaviour for $x$ close to $1$ to be $approx -{2 over x-1}$.
answered Dec 9 '18 at 6:56
copper.hat
126k559159
Write ${1 over 1-x} - {1 over log x} = {log x +x-1over (1-x) log x}$.
The Taylor expansions are
$log x +x-1 = 2(x-1)+cdots$ and $(1-x) log x = -(x-1)^2+ cdots $.
Hence we expect the behaviour for $x$ close to $1$ to be $approx -{2 over x-1}$.
answered Dec 9 '18 at 6:56
copper.hat
126k559159
answered Dec 9 '18 at 6:56
copper.hat
126k559159
answered Dec 9 '18 at 6:56
copper.hat
126k559159
answered Dec 9 '18 at 6:56
copper.hat
126k559159
126k559159
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
add a comment |
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Actually I couldn't understand what you mean,can you write as more detailed?
– Atakan
Dec 9 '18 at 7:06
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
Are you familiar with Taylor expansion?
– copper.hat
Dec 9 '18 at 7:10
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
no,I'm not familiar :(
– Atakan
Dec 9 '18 at 7:11
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
Certain functions can be expanded around a point in the form $f(x) = sum_{k ge 0} f^{(k)}(x_o) (x-x_0)^k$. (en.wikipedia.org/wiki/Taylor_series.) This expansion can be used to approximate the local behaviour of $f$. In this case, I am approximating the numerator and denominator by Taylor series, and the leading (non zero) terms dictate the quotient's behaviour.
– copper.hat
Dec 9 '18 at 7:15
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
okay, I'm trying to understand it
– Atakan
Dec 9 '18 at 7:24
add a comment |
I want to rewrite the limit slightly:
$$lim_{xto1} left(frac{1}{1-x} - frac{1}{ln(x)}right) = -lim_{xto1} left(frac{1}{ln(x)} + frac{1}{x-1} right). $$
Note that now our two summands have the same sign when $x<1$ and when $x>1$. The limit from the left goes to $(-infty)+ (-infty) = (-infty)$ and the limit from the right goes to $infty + infty = infty$. Thus, the limit does not exist.
No indefinite forms needed!
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
add a comment |
I want to rewrite the limit slightly:
$$lim_{xto1} left(frac{1}{1-x} - frac{1}{ln(x)}right) = -lim_{xto1} left(frac{1}{ln(x)} + frac{1}{x-1} right). $$
Note that now our two summands have the same sign when $x<1$ and when $x>1$. The limit from the left goes to $(-infty)+ (-infty) = (-infty)$ and the limit from the right goes to $infty + infty = infty$. Thus, the limit does not exist.
No indefinite forms needed!
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
add a comment |
I want to rewrite the limit slightly:
$$lim_{xto1} left(frac{1}{1-x} - frac{1}{ln(x)}right) = -lim_{xto1} left(frac{1}{ln(x)} + frac{1}{x-1} right). $$
Note that now our two summands have the same sign when $x<1$ and when $x>1$. The limit from the left goes to $(-infty)+ (-infty) = (-infty)$ and the limit from the right goes to $infty + infty = infty$. Thus, the limit does not exist.
No indefinite forms needed!
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
I want to rewrite the limit slightly:
$$lim_{xto1} left(frac{1}{1-x} - frac{1}{ln(x)}right) = -lim_{xto1} left(frac{1}{ln(x)} + frac{1}{x-1} right). $$
Note that now our two summands have the same sign when $x<1$ and when $x>1$. The limit from the left goes to $(-infty)+ (-infty) = (-infty)$ and the limit from the right goes to $infty + infty = infty$. Thus, the limit does not exist.
No indefinite forms needed!
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
answered Dec 9 '18 at 6:57
Santana Afton
2,5742629
2,5742629
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
add a comment |
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
Yes,you solved it with small changes but it's looking true ,thank you for your solving :)
– Atakan
Dec 9 '18 at 7:08
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
That's the best solution in my opinion! Simple and clever.
– gimusi
Dec 9 '18 at 14:49
add a comment |
I guess the first term should have been $frac{1}{x-1}$ rather than $frac{1}{1-x}$. In the former case the substitution $x=e^t$ reduces the problem to the evaluation of
$$ lim_{tto 0}left(frac{1}{e^t-1}-frac{1}{t}right)=-lim_{t to 0}frac{e^t-1-t}{t(e^t-1)}=-lim_{tto 0}frac{frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=color{red}{-frac{1}{2}}. $$
The Maclaurin series of $e^t$ is either a trivial consequence of the series definition of $e^t$ or a simple consequence of the dominated convergence theorem / integration by parts:
$$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
add a comment |
I guess the first term should have been $frac{1}{x-1}$ rather than $frac{1}{1-x}$. In the former case the substitution $x=e^t$ reduces the problem to the evaluation of
$$ lim_{tto 0}left(frac{1}{e^t-1}-frac{1}{t}right)=-lim_{t to 0}frac{e^t-1-t}{t(e^t-1)}=-lim_{tto 0}frac{frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=color{red}{-frac{1}{2}}. $$
The Maclaurin series of $e^t$ is either a trivial consequence of the series definition of $e^t$ or a simple consequence of the dominated convergence theorem / integration by parts:
$$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
add a comment |
I guess the first term should have been $frac{1}{x-1}$ rather than $frac{1}{1-x}$. In the former case the substitution $x=e^t$ reduces the problem to the evaluation of
$$ lim_{tto 0}left(frac{1}{e^t-1}-frac{1}{t}right)=-lim_{t to 0}frac{e^t-1-t}{t(e^t-1)}=-lim_{tto 0}frac{frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=color{red}{-frac{1}{2}}. $$
The Maclaurin series of $e^t$ is either a trivial consequence of the series definition of $e^t$ or a simple consequence of the dominated convergence theorem / integration by parts:
$$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
I guess the first term should have been $frac{1}{x-1}$ rather than $frac{1}{1-x}$. In the former case the substitution $x=e^t$ reduces the problem to the evaluation of
$$ lim_{tto 0}left(frac{1}{e^t-1}-frac{1}{t}right)=-lim_{t to 0}frac{e^t-1-t}{t(e^t-1)}=-lim_{tto 0}frac{frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=color{red}{-frac{1}{2}}. $$
The Maclaurin series of $e^t$ is either a trivial consequence of the series definition of $e^t$ or a simple consequence of the dominated convergence theorem / integration by parts:
$$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
answered Dec 9 '18 at 11:45
Jack D'Aurizio
287k33280657
287k33280657
add a comment |
add a comment |
We have that by $y=x-1 to 0$
$$lim_{xto1} left(dfrac{1}{1-x}-dfrac{1}{ln x}right) =lim_{yto0} left(-dfrac{1}{y}-dfrac{1}{ln (1+y)}right)$$
and by standard limits since $frac{ln(1+y)}yto 1$
$$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$
answered Dec 9 '18 at 8:12
gimusi
1
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
add a comment |
We have that by $y=x-1 to 0$
$$lim_{xto1} left(dfrac{1}{1-x}-dfrac{1}{ln x}right) =lim_{yto0} left(-dfrac{1}{y}-dfrac{1}{ln (1+y)}right)$$
and by standard limits since $frac{ln(1+y)}yto 1$
$$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$
answered Dec 9 '18 at 8:12
gimusi
1
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
add a comment |
We have that by $y=x-1 to 0$
$$lim_{xto1} left(dfrac{1}{1-x}-dfrac{1}{ln x}right) =lim_{yto0} left(-dfrac{1}{y}-dfrac{1}{ln (1+y)}right)$$
and by standard limits since $frac{ln(1+y)}yto 1$
$$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$
answered Dec 9 '18 at 8:12
gimusi
1
We have that by $y=x-1 to 0$
$$lim_{xto1} left(dfrac{1}{1-x}-dfrac{1}{ln x}right) =lim_{yto0} left(-dfrac{1}{y}-dfrac{1}{ln (1+y)}right)$$
and by standard limits since $frac{ln(1+y)}yto 1$
$$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$
answered Dec 9 '18 at 8:12
gimusi
1
edited Dec 9 '18 at 12:34
answered Dec 9 '18 at 8:12
gimusi
1
answered Dec 9 '18 at 8:12
gimusi
1
answered Dec 9 '18 at 8:12
gimusi
1
1
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
add a comment |
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
$+infty$ when $yto0^-$
– Shubham Johri
Dec 9 '18 at 12:28
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
@ShubhamJohri Yes indeed I lost a minus sign in the last step and reverse the result! Thanks so much for pointing out that!
– gimusi
Dec 9 '18 at 12:33
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
You're welcome ;)
– Shubham Johri
Dec 9 '18 at 12:35
add a comment |
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@copper.hat I put a picture link into sentence.You can look at it
– Atakan
Dec 9 '18 at 6:12
Are you familiar with L'Hôpital's rule? (The rule can be repeated as necessary.)
– copper.hat
Dec 9 '18 at 6:25
1
@Atakan: Limit does not exist check by $xto 1^-$ and $xto1^+$.
– Yadati Kiran
Dec 9 '18 at 6:28
1
@AniruddhVenkatesan: Yeah. as $xto1^-$ we have $+infty$ and $xto1^+$ we have $-infty$
– Yadati Kiran
Dec 9 '18 at 6:31
1
@Atakan: The two sided limit does not exist.
– Yadati Kiran
Dec 9 '18 at 6:34