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Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to prove that $limlimits_{xto0}frac{sin x}x=1$?L'Hopital's Rule with $lim limits_{x to infty}frac{2^x}{e^left(x^2right)}$How do I find $limlimits_{x to 0} xcot(6x) $ without using L'hopitals rule?Prove: $limlimits_{ntoinfty}frac{3^nsin{n^3}}{3^{n-1}+3n}=3$Evaluate $limlimits_{xtoinfty}x(frac{pi}{2}-arctan(x))$ without using L'HôpitalEvaluate $limlimits_{x to infty} sin(frac{1}{x})^x$Evaluate $limlimits frac{x-sinsincdotssin x}{x^3}.$Evaluate the following integral $limlimits_{xto 0}frac{1}{x}int_{0}^{x}sin^{2}left(frac{1}{u}right)du$Why doesn't L'hopitals Rule work for $limlimits_{x to infty} dfrac{x+ sin x}{x+ 2 sin x}$Compute $limlimits_{nto infty} n sqrt{2^{n-1}} cos^{n-1} alpha$Evaluate $limlimits_{xtoinfty}int^{2x}_{x}frac{1}{t}dt$












0












$begingroup$


Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$



My trial



As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,



begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}



This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?






calculus algebra-precalculus limits







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$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/75130/…
    $endgroup$
    – Robert Z
    Mar 26 at 10:23










  • $begingroup$
    @Robert Z: Thanks a lot, I got that hint!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:26
















0












$begingroup$


Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$



My trial



As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,



begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}



This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?






calculus algebra-precalculus limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/75130/…
    $endgroup$
    – Robert Z
    Mar 26 at 10:23










  • $begingroup$
    @Robert Z: Thanks a lot, I got that hint!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:26














0












0








0


2



$begingroup$


Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$



My trial



As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,



begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}



This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?






calculus algebra-precalculus limits







share|cite|improve this question









$endgroup$




Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$



My trial



As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,



begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}



This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?






calculus algebra-precalculus limits




calculus algebra-precalculus limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 26 at 10:21









Omojola MichealOmojola Micheal

2,082424




2,082424












  • $begingroup$
    See math.stackexchange.com/questions/75130/…
    $endgroup$
    – Robert Z
    Mar 26 at 10:23










  • $begingroup$
    @Robert Z: Thanks a lot, I got that hint!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:26


















  • $begingroup$
    See math.stackexchange.com/questions/75130/…
    $endgroup$
    – Robert Z
    Mar 26 at 10:23










  • $begingroup$
    @Robert Z: Thanks a lot, I got that hint!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:26
















$begingroup$
See math.stackexchange.com/questions/75130/…
$endgroup$
– Robert Z
Mar 26 at 10:23




$begingroup$
See math.stackexchange.com/questions/75130/…
$endgroup$
– Robert Z
Mar 26 at 10:23












$begingroup$
@Robert Z: Thanks a lot, I got that hint!
$endgroup$
– Omojola Micheal
Mar 26 at 10:26




$begingroup$
@Robert Z: Thanks a lot, I got that hint!
$endgroup$
– Omojola Micheal
Mar 26 at 10:26










2 Answers
2






active

oldest

votes


















5












$begingroup$

$ frac{x}{sin x} to 1$ as $x to 0$,



hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, thanks Fred!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:25



















2












$begingroup$

Hint:




  1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


  2. $ln$ is continuous.


  3. $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    $ frac{x}{sin x} to 1$ as $x to 0$,



    hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh, thanks Fred!
      $endgroup$
      – Omojola Micheal
      Mar 26 at 10:25
















    5












    $begingroup$

    $ frac{x}{sin x} to 1$ as $x to 0$,



    hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh, thanks Fred!
      $endgroup$
      – Omojola Micheal
      Mar 26 at 10:25














    5












    5








    5





    $begingroup$

    $ frac{x}{sin x} to 1$ as $x to 0$,



    hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$






    share|cite|improve this answer









    $endgroup$



    $ frac{x}{sin x} to 1$ as $x to 0$,



    hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 26 at 10:24









    FredFred

    48.6k11849




    48.6k11849












    • $begingroup$
      Oh, thanks Fred!
      $endgroup$
      – Omojola Micheal
      Mar 26 at 10:25


















    • $begingroup$
      Oh, thanks Fred!
      $endgroup$
      – Omojola Micheal
      Mar 26 at 10:25
















    $begingroup$
    Oh, thanks Fred!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:25




    $begingroup$
    Oh, thanks Fred!
    $endgroup$
    – Omojola Micheal
    Mar 26 at 10:25











    2












    $begingroup$

    Hint:




    1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


    2. $ln$ is continuous.


    3. $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:




      1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


      2. $ln$ is continuous.


      3. $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:




        1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


        2. $ln$ is continuous.


        3. $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.






        share|cite|improve this answer









        $endgroup$



        Hint:




        1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


        2. $ln$ is continuous.


        3. $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 26 at 10:26









        5xum5xum

        92.9k395162




        92.9k395162






























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