Help understanding the steps of a solved limit












2














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 .










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.

























    2














    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 .










    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.























      2












      2








      2







      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 .










      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






      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 Dec 22 at 15:09





















      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.









      asked Dec 22 at 14:08









      El Bryan

      134




      134




      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.





      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.






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






















          2 Answers
          2






          active

          oldest

          votes


















          4














          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
            Dec 22 at 14:42










          • 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
            Dec 22 at 14:47






          • 1




            Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
            – El Bryan
            Dec 22 at 15:11










          • @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
            Dec 22 at 15:15



















          2














          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























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });






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










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049490%2fhelp-understanding-the-steps-of-a-solved-limit%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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
              Dec 22 at 14:42










            • 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
              Dec 22 at 14:47






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              Dec 22 at 15:11










            • @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
              Dec 22 at 15:15
















            4














            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
              Dec 22 at 14:42










            • 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
              Dec 22 at 14:47






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              Dec 22 at 15:11










            • @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
              Dec 22 at 15:15














            4












            4








            4






            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 Dec 22 at 14:54









            Bernard

            118k639112




            118k639112










            answered Dec 22 at 14:30









            Ankit Kumar

            1,593119




            1,593119












            • You used taylor? also what are typos (I'm new here)
              – El Bryan
              Dec 22 at 14:42










            • 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
              Dec 22 at 14:47






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              Dec 22 at 15:11










            • @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
              Dec 22 at 15:15


















            • You used taylor? also what are typos (I'm new here)
              – El Bryan
              Dec 22 at 14:42










            • 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
              Dec 22 at 14:47






            • 1




              Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
              – El Bryan
              Dec 22 at 15:11










            • @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
              Dec 22 at 15:15
















            You used taylor? also what are typos (I'm new here)
            – El Bryan
            Dec 22 at 14:42




            You used taylor? also what are typos (I'm new here)
            – El Bryan
            Dec 22 at 14:42












            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
            Dec 22 at 14:47




            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
            Dec 22 at 14:47




            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
            Dec 22 at 15:11




            Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
            – El Bryan
            Dec 22 at 15:11












            @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
            Dec 22 at 15:15




            @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
            Dec 22 at 15:15











            2














            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




























              2














              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


























                2












                2








                2






                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 Dec 22 at 14:55

























                answered Dec 22 at 14:29









                roman

                1,92921221




                1,92921221






















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










                    draft saved

                    draft discarded


















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













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












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
















                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049490%2fhelp-understanding-the-steps-of-a-solved-limit%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    數位音樂下載

                    When can things happen in Etherscan, such as the picture below?

                    格利澤436b