Help understanding the steps of a solved limit












1














A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:



begin{align}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
&=lim_{x rightarrow +∞}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\
&=
lim_{x rightarrow +∞}
frac{2x}{(6x^3)}= 0
end{align}



The part that I don't understand is why the limit is equal to:



begin{equation*}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
end{equation*}



The only thing I found out so far is that:



begin{equation*}
sin (x) - x ∼ x^3/6
end{equation*}



So:
begin{equation*}
sin (1/x) - (1/x)∼ 1/(6x^3)
end{equation*}



For the rest, I have no idea .






limits asymptotics







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    1














    A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:



    begin{align}
    lim_{x rightarrow +∞}
    frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
    &=lim_{x rightarrow +∞}
    frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
    \
    &=
    lim_{x rightarrow +∞}
    frac{2x}{(6x^3)}= 0
    end{align}



    The part that I don't understand is why the limit is equal to:



    begin{equation*}
    frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
    end{equation*}



    The only thing I found out so far is that:



    begin{equation*}
    sin (x) - x ∼ x^3/6
    end{equation*}



    So:
    begin{equation*}
    sin (1/x) - (1/x)∼ 1/(6x^3)
    end{equation*}



    For the rest, I have no idea .






    limits asymptotics







    share|cite|improve this question









    New contributor




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























      1












      1








      1







      A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:



      begin{align}
      lim_{x rightarrow +∞}
      frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
      &=lim_{x rightarrow +∞}
      frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
      \
      &=
      lim_{x rightarrow +∞}
      frac{2x}{(6x^3)}= 0
      end{align}



      The part that I don't understand is why the limit is equal to:



      begin{equation*}
      frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
      end{equation*}



      The only thing I found out so far is that:



      begin{equation*}
      sin (x) - x ∼ x^3/6
      end{equation*}



      So:
      begin{equation*}
      sin (1/x) - (1/x)∼ 1/(6x^3)
      end{equation*}



      For the rest, I have no idea .






      limits asymptotics







      share|cite|improve this question









      New contributor




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











      A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:



      begin{align}
      lim_{x rightarrow +∞}
      frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
      &=lim_{x rightarrow +∞}
      frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
      \
      &=
      lim_{x rightarrow +∞}
      frac{2x}{(6x^3)}= 0
      end{align}



      The part that I don't understand is why the limit is equal to:



      begin{equation*}
      frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
      end{equation*}



      The only thing I found out so far is that:



      begin{equation*}
      sin (x) - x ∼ x^3/6
      end{equation*}



      So:
      begin{equation*}
      sin (1/x) - (1/x)∼ 1/(6x^3)
      end{equation*}



      For the rest, I have no idea .






      limits asymptotics




      limits asymptotics






      share|cite|improve this question









      New contributor




      El Bryan 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




      El Bryan 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





















      New contributor




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      asked 2 hours ago









      El Bryan

      84




      84




      New contributor




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      New contributor





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






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          2 Answers
          2






          active

          oldest

          votes


















          3














          It isn't exactly correct.



          $$sin(x)=x-x^3/3!+x^5/5!-cdots$$
          This can be obtained from Taylor's expansion.



          For $log(1+x),$
          $$log(1+x)=intfrac{1}{1+x}dx$$
          $$=int (1-x+x^2-x^3+cdots)dx$$
          $$log(1+x)=x-x^2/2+cdots$$
          So, your limit,
          begin{equation*}
          lim_{x rightarrow +∞}
          frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
          end{equation*}



          begin{equation*}
          lim_{x rightarrow +∞}
          frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
          end{equation*}
          Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
          begin{equation*}
          lim_{x rightarrow +∞}
          frac{2x}{6x^3}= 0
          end{equation*}
          Your solution has a lot of typos






          share|cite|improve this answer























          • You used taylor? also what are typos (I'm new here)
            – El Bryan
            1 hour ago










          • I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
            – Ankit Kumar
            1 hour ago






          • 1




            Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
            – El Bryan
            1 hour ago










          • @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
            – Ankit Kumar
            1 hour ago



















          1














          First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
          $$
          lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
          $$



          By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
          $$
          sin t = t - {t^3over 3!} + {t^5over 5!} - dots
          $$



          At the same time for $log(1+t)$:
          $$
          log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
          $$



          So if you apply this to your limit you'll observe the desired result:
          $$
          begin{align}
          lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
          &= lim_{tto 0} frac{2t^3}{6t} = \
          &= lim_{tto 0} frac{2t^2}{6} = 0
          end{align}
          $$






          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









            3














            It isn't exactly correct.



            $$sin(x)=x-x^3/3!+x^5/5!-cdots$$
            This can be obtained from Taylor's expansion.



            For $log(1+x),$
            $$log(1+x)=intfrac{1}{1+x}dx$$
            $$=int (1-x+x^2-x^3+cdots)dx$$
            $$log(1+x)=x-x^2/2+cdots$$
            So, your limit,
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
            end{equation*}



            begin{equation*}
            lim_{x rightarrow +∞}
            frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
            end{equation*}
            Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{2x}{6x^3}= 0
            end{equation*}
            Your solution has a lot of typos






            share|cite|improve this answer























            • You used taylor? also what are typos (I'm new here)
              – El Bryan
              1 hour ago










            • I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
              – Ankit Kumar
              1 hour ago






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              1 hour ago










            • @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
              – Ankit Kumar
              1 hour ago
















            3














            It isn't exactly correct.



            $$sin(x)=x-x^3/3!+x^5/5!-cdots$$
            This can be obtained from Taylor's expansion.



            For $log(1+x),$
            $$log(1+x)=intfrac{1}{1+x}dx$$
            $$=int (1-x+x^2-x^3+cdots)dx$$
            $$log(1+x)=x-x^2/2+cdots$$
            So, your limit,
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
            end{equation*}



            begin{equation*}
            lim_{x rightarrow +∞}
            frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
            end{equation*}
            Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{2x}{6x^3}= 0
            end{equation*}
            Your solution has a lot of typos






            share|cite|improve this answer























            • You used taylor? also what are typos (I'm new here)
              – El Bryan
              1 hour ago










            • I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
              – Ankit Kumar
              1 hour ago






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              1 hour ago










            • @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
              – Ankit Kumar
              1 hour ago














            3












            3








            3






            It isn't exactly correct.



            $$sin(x)=x-x^3/3!+x^5/5!-cdots$$
            This can be obtained from Taylor's expansion.



            For $log(1+x),$
            $$log(1+x)=intfrac{1}{1+x}dx$$
            $$=int (1-x+x^2-x^3+cdots)dx$$
            $$log(1+x)=x-x^2/2+cdots$$
            So, your limit,
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
            end{equation*}



            begin{equation*}
            lim_{x rightarrow +∞}
            frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
            end{equation*}
            Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{2x}{6x^3}= 0
            end{equation*}
            Your solution has a lot of typos






            share|cite|improve this answer














            It isn't exactly correct.



            $$sin(x)=x-x^3/3!+x^5/5!-cdots$$
            This can be obtained from Taylor's expansion.



            For $log(1+x),$
            $$log(1+x)=intfrac{1}{1+x}dx$$
            $$=int (1-x+x^2-x^3+cdots)dx$$
            $$log(1+x)=x-x^2/2+cdots$$
            So, your limit,
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
            end{equation*}



            begin{equation*}
            lim_{x rightarrow +∞}
            frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
            end{equation*}
            Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
            begin{equation*}
            lim_{x rightarrow +∞}
            frac{2x}{6x^3}= 0
            end{equation*}
            Your solution has a lot of typos







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago









            Bernard

            117k637111




            117k637111










            answered 2 hours ago









            Ankit Kumar

            1,04217




            1,04217












            • You used taylor? also what are typos (I'm new here)
              – El Bryan
              1 hour ago










            • I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
              – Ankit Kumar
              1 hour ago






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              1 hour ago










            • @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
              – Ankit Kumar
              1 hour ago


















            • You used taylor? also what are typos (I'm new here)
              – El Bryan
              1 hour ago










            • I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
              – Ankit Kumar
              1 hour ago






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              1 hour ago










            • @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
              – Ankit Kumar
              1 hour ago
















            You used taylor? also what are typos (I'm new here)
            – El Bryan
            1 hour ago




            You used taylor? also what are typos (I'm new here)
            – El Bryan
            1 hour ago












            I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
            – Ankit Kumar
            1 hour ago




            I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
            – Ankit Kumar
            1 hour ago




            1




            1




            Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
            – El Bryan
            1 hour ago




            Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
            – El Bryan
            1 hour ago












            @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
            – Ankit Kumar
            1 hour ago




            @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
            – Ankit Kumar
            1 hour ago











            1














            First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
            $$
            lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
            $$



            By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
            $$
            sin t = t - {t^3over 3!} + {t^5over 5!} - dots
            $$



            At the same time for $log(1+t)$:
            $$
            log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
            $$



            So if you apply this to your limit you'll observe the desired result:
            $$
            begin{align}
            lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
            &= lim_{tto 0} frac{2t^3}{6t} = \
            &= lim_{tto 0} frac{2t^2}{6} = 0
            end{align}
            $$






            share|cite|improve this answer




























              1














              First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
              $$
              lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
              $$



              By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
              $$
              sin t = t - {t^3over 3!} + {t^5over 5!} - dots
              $$



              At the same time for $log(1+t)$:
              $$
              log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
              $$



              So if you apply this to your limit you'll observe the desired result:
              $$
              begin{align}
              lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
              &= lim_{tto 0} frac{2t^3}{6t} = \
              &= lim_{tto 0} frac{2t^2}{6} = 0
              end{align}
              $$






              share|cite|improve this answer


























                1












                1








                1






                First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
                $$
                lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
                $$



                By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
                $$
                sin t = t - {t^3over 3!} + {t^5over 5!} - dots
                $$



                At the same time for $log(1+t)$:
                $$
                log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
                $$



                So if you apply this to your limit you'll observe the desired result:
                $$
                begin{align}
                lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
                &= lim_{tto 0} frac{2t^3}{6t} = \
                &= lim_{tto 0} frac{2t^2}{6} = 0
                end{align}
                $$






                share|cite|improve this answer














                First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
                $$
                lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
                $$



                By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
                $$
                sin t = t - {t^3over 3!} + {t^5over 5!} - dots
                $$



                At the same time for $log(1+t)$:
                $$
                log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
                $$



                So if you apply this to your limit you'll observe the desired result:
                $$
                begin{align}
                lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
                &= lim_{tto 0} frac{2t^3}{6t} = \
                &= lim_{tto 0} frac{2t^2}{6} = 0
                end{align}
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 2 hours ago









                roman

                1,67511120




                1,67511120






















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