Given $lim_{x rightarrow a}f(x)+frac1{left|f(x)right|}=0$, what is $lim_{x rightarrow a}f(x)$?

Multi tool use
Multi tool use












4












$begingroup$


Another question I am struggling with:



Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




Thank you all for your answers!!



But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






calculus limits







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Another question I am struggling with:



    Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




    Thank you all for your answers!!



    But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






    calculus limits







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Another question I am struggling with:



      Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




      Thank you all for your answers!!



      But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






      calculus limits







      share|cite|improve this question











      $endgroup$




      Another question I am struggling with:



      Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




      Thank you all for your answers!!



      But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






      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 Jan 13 at 13:30









      Did

      248k23225464




      248k23225464










      asked Nov 30 '13 at 11:27









      user83081user83081

      41428




      41428






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47



















          4












          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47











          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
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f586700%2fgiven-lim-x-rightarrow-afx-frac1-leftfx-right-0-what-is-lim-x%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









          6












          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          6












          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47














          6












          6








          6





          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$



          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '13 at 12:02

























          answered Nov 30 '13 at 11:38









          Paramanand SinghParamanand Singh

          50.8k557168




          50.8k557168












          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47


















          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47




          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47











          4












          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          4












          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47














          4












          4








          4





          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$



          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '13 at 13:40

























          answered Nov 30 '13 at 12:46









          Christian BlatterChristian Blatter

          175k8115327




          175k8115327












          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47


















          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47




          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47


















          draft saved

          draft discarded




















































          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f586700%2fgiven-lim-x-rightarrow-afx-frac1-leftfx-right-0-what-is-lim-x%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







          Iy2,XXT2H3i50v5i6 K9V3 LwfGNugC3eRF8NkVy,xLrmuamkJL6MFBNiI0dfOkx3MRej0W87Mu9apNtsmXQ IgTbaTX6dvlrQ,azkgdx
          -
          lhGZ IhxM23 iOiozf,I7k vF8neeMML6k D,1lVFpmSm lAqYBjM,Kbq2QCJ4OrNO,9Pk,57wZH

          Popular posts from this blog

          Human spaceflight

          Image
          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
          Read more

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
          Read more

          張江高科駅

          Image
          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
          Read more