How to develop a solution in limit without L'Hopital?












2















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







share|cite|improve this question
























  • @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
















2















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







share|cite|improve this question
























  • @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














2












2








2








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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 7:20

























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


















  • @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










5 Answers
5






active

oldest

votes


















4














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 $mto0^+implies xto1^-$.



enter image description here






share|cite|improve this answer























  • 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



















2














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 $approx -{2 over x-1}$.






share|cite|improve this answer





















  • 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



















2














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!






share|cite|improve this answer





















  • 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





















2














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:
$$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$






share|cite|improve this answer





























    2














    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 $frac{ln(1+y)}yto 1$



    $$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$






    share|cite|improve this answer























    • $+infty$ 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











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    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    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 $mto0^+implies xto1^-$.



    enter image description here






    share|cite|improve this answer























    • 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
















    4














    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 $mto0^+implies xto1^-$.



    enter image description here






    share|cite|improve this answer























    • 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














    4












    4








    4






    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 $mto0^+implies xto1^-$.



    enter image description here






    share|cite|improve this answer














    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 $mto0^+implies xto1^-$.



    enter image description here







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 12:26

























    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


















    • 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











    2














    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 $approx -{2 over x-1}$.






    share|cite|improve this answer





















    • 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
















    2














    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 $approx -{2 over x-1}$.






    share|cite|improve this answer





















    • 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














    2












    2








    2






    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 $approx -{2 over x-1}$.






    share|cite|improve this answer












    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 $approx -{2 over x-1}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • 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











    2














    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!






    share|cite|improve this answer





















    • 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


















    2














    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!






    share|cite|improve this answer





















    • 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
















    2












    2








    2






    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!






    share|cite|improve this answer












    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!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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




















    • 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













    2














    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:
    $$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$






    share|cite|improve this answer


























      2














      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:
      $$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$






      share|cite|improve this answer
























        2












        2








        2






        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:
        $$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$






        share|cite|improve this answer












        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:
        $$ e^t-1-t = t^2int_{0}^{1}(1-x)e^{tx},dx to t^2 int_{0}^{1}(1-x),dx = frac{t^2}{2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 11:45









        Jack D'Aurizio

        287k33280657




        287k33280657























            2














            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 $frac{ln(1+y)}yto 1$



            $$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$






            share|cite|improve this answer























            • $+infty$ 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
















            2














            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 $frac{ln(1+y)}yto 1$



            $$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$






            share|cite|improve this answer























            • $+infty$ 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














            2












            2








            2






            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 $frac{ln(1+y)}yto 1$



            $$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$






            share|cite|improve this answer














            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 $frac{ln(1+y)}yto 1$



            $$-dfrac{1}{y}-dfrac{1}{ln (1+y)}=-dfrac{ln(1+y)+y}{yln (1+y)}=-dfrac{frac{ln(1+y)}y+1}{ln (1+y)}to begin{cases}-inftyquad yto 0^+\\+inftyquad yto 0^- end{cases}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 12:34

























            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


















            • $+infty$ 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


















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