How can i claim a one-sided limit doesnt exist?












1












$begingroup$


I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$



Now, I tried using L'Hôpital's rule and it doesn't lead anywhere.



Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.



However, I cannot prove it. What can I do?






real-analysis limits proof-writing logarithms limits-without-lhopital







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  • 1




    $begingroup$
    Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
    $endgroup$
    – Yves Daoust
    1 hour ago










  • $begingroup$
    We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
    $endgroup$
    – sdds
    1 hour ago


















1












$begingroup$


I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$



Now, I tried using L'Hôpital's rule and it doesn't lead anywhere.



Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.



However, I cannot prove it. What can I do?






real-analysis limits proof-writing logarithms limits-without-lhopital







share|cite|improve this question









New contributor




sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
    $endgroup$
    – Yves Daoust
    1 hour ago










  • $begingroup$
    We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
    $endgroup$
    – sdds
    1 hour ago
















1












1








1





$begingroup$


I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$



Now, I tried using L'Hôpital's rule and it doesn't lead anywhere.



Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.



However, I cannot prove it. What can I do?






real-analysis limits proof-writing logarithms limits-without-lhopital







share|cite|improve this question









New contributor




sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$



Now, I tried using L'Hôpital's rule and it doesn't lead anywhere.



Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.



However, I cannot prove it. What can I do?






real-analysis limits proof-writing logarithms limits-without-lhopital




real-analysis limits proof-writing logarithms limits-without-lhopital






share|cite|improve this question









New contributor




sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Michael Rozenberg

100k1591192




100k1591192






New contributor




sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









sddssdds

61




61




New contributor




sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






sdds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
    $endgroup$
    – Yves Daoust
    1 hour ago










  • $begingroup$
    We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
    $endgroup$
    – sdds
    1 hour ago
















  • 1




    $begingroup$
    Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
    $endgroup$
    – Yves Daoust
    1 hour ago










  • $begingroup$
    We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
    $endgroup$
    – sdds
    1 hour ago










1




1




$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
1 hour ago




$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
1 hour ago












$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
1 hour ago






$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
1 hour ago












3 Answers
3






active

oldest

votes


















5












$begingroup$

$$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Hint:



    Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.



    To do so, rewrite, for $x>0$,
    $$
    frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, this conversion never occurred to me
      $endgroup$
      – sdds
      1 hour ago










    • $begingroup$
      @sdds You're welcome!
      $endgroup$
      – Clement C.
      1 hour ago



















    0












    $begingroup$

    It is known that



    $$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$



    Then



    $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It's true, but how is that different from the current two answers?
      $endgroup$
      – Clement C.
      1 hour ago











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    $$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      $$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        $$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$






        share|cite|improve this answer









        $endgroup$



        $$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Michael RozenbergMichael Rozenberg

        100k1591192




        100k1591192























            4












            $begingroup$

            Hint:



            Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.



            To do so, rewrite, for $x>0$,
            $$
            frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you, this conversion never occurred to me
              $endgroup$
              – sdds
              1 hour ago










            • $begingroup$
              @sdds You're welcome!
              $endgroup$
              – Clement C.
              1 hour ago
















            4












            $begingroup$

            Hint:



            Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.



            To do so, rewrite, for $x>0$,
            $$
            frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you, this conversion never occurred to me
              $endgroup$
              – sdds
              1 hour ago










            • $begingroup$
              @sdds You're welcome!
              $endgroup$
              – Clement C.
              1 hour ago














            4












            4








            4





            $begingroup$

            Hint:



            Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.



            To do so, rewrite, for $x>0$,
            $$
            frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
            $$






            share|cite|improve this answer









            $endgroup$



            Hint:



            Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.



            To do so, rewrite, for $x>0$,
            $$
            frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Clement C.Clement C.

            50.1k33888




            50.1k33888












            • $begingroup$
              Thank you, this conversion never occurred to me
              $endgroup$
              – sdds
              1 hour ago










            • $begingroup$
              @sdds You're welcome!
              $endgroup$
              – Clement C.
              1 hour ago


















            • $begingroup$
              Thank you, this conversion never occurred to me
              $endgroup$
              – sdds
              1 hour ago










            • $begingroup$
              @sdds You're welcome!
              $endgroup$
              – Clement C.
              1 hour ago
















            $begingroup$
            Thank you, this conversion never occurred to me
            $endgroup$
            – sdds
            1 hour ago




            $begingroup$
            Thank you, this conversion never occurred to me
            $endgroup$
            – sdds
            1 hour ago












            $begingroup$
            @sdds You're welcome!
            $endgroup$
            – Clement C.
            1 hour ago




            $begingroup$
            @sdds You're welcome!
            $endgroup$
            – Clement C.
            1 hour ago











            0












            $begingroup$

            It is known that



            $$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$



            Then



            $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              It's true, but how is that different from the current two answers?
              $endgroup$
              – Clement C.
              1 hour ago
















            0












            $begingroup$

            It is known that



            $$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$



            Then



            $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              It's true, but how is that different from the current two answers?
              $endgroup$
              – Clement C.
              1 hour ago














            0












            0








            0





            $begingroup$

            It is known that



            $$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$



            Then



            $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$






            share|cite|improve this answer









            $endgroup$



            It is known that



            $$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$



            Then



            $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Yves DaoustYves Daoust

            126k672225




            126k672225












            • $begingroup$
              It's true, but how is that different from the current two answers?
              $endgroup$
              – Clement C.
              1 hour ago


















            • $begingroup$
              It's true, but how is that different from the current two answers?
              $endgroup$
              – Clement C.
              1 hour ago
















            $begingroup$
            It's true, but how is that different from the current two answers?
            $endgroup$
            – Clement C.
            1 hour ago




            $begingroup$
            It's true, but how is that different from the current two answers?
            $endgroup$
            – Clement C.
            1 hour ago










            sdds is a new contributor. Be nice, and check out our Code of Conduct.










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            sdds is a new contributor. Be nice, and check out our Code of Conduct.












            sdds is a new contributor. Be nice, and check out our Code of Conduct.
















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