Using Partial Limit












2












$begingroup$


$$lim_{x to 0} cos(pi/2cos(x))/x^2$$



I tried to evaluate the limit this way,



$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$



$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$



Now apply L'Hospital's Rule twice,



$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$



So,this way the answer is zero.



Can you please explain where am I doing wrong?



I will be thankful for help!






calculus limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
    $endgroup$
    – Navneet Kumar
    2 hours ago












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    2 hours ago










  • $begingroup$
    "The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago
















2












$begingroup$


$$lim_{x to 0} cos(pi/2cos(x))/x^2$$



I tried to evaluate the limit this way,



$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$



$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$



Now apply L'Hospital's Rule twice,



$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$



So,this way the answer is zero.



Can you please explain where am I doing wrong?



I will be thankful for help!






calculus limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
    $endgroup$
    – Navneet Kumar
    2 hours ago












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    2 hours ago










  • $begingroup$
    "The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago














2












2








2





$begingroup$


$$lim_{x to 0} cos(pi/2cos(x))/x^2$$



I tried to evaluate the limit this way,



$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$



$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$



Now apply L'Hospital's Rule twice,



$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$



So,this way the answer is zero.



Can you please explain where am I doing wrong?



I will be thankful for help!






calculus limits







share|cite|improve this question











$endgroup$




$$lim_{x to 0} cos(pi/2cos(x))/x^2$$



I tried to evaluate the limit this way,



$$lim_{x to 0} cos(pi/2cdot1)/x^2$$ since $cos0=1$



$$lim_{x to 0} cos(pi/2cdot1)/x^2=lim_{x to 0} 0/x^2$$



Now apply L'Hospital's Rule twice,



$$lim_{x to 0} 0/2(x)=lim_{x to 0} 0/2=0$$



So,this way the answer is zero.



Can you please explain where am I doing wrong?



I will be thankful for help!






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 3 hours ago









Larry

2,1162826




2,1162826










asked 3 hours ago









Navneet KumarNavneet Kumar

3881416




3881416








  • 2




    $begingroup$
    You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
    $endgroup$
    – Navneet Kumar
    2 hours ago












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    2 hours ago










  • $begingroup$
    "The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago














  • 2




    $begingroup$
    You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
    $endgroup$
    – Navneet Kumar
    2 hours ago












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    2 hours ago










  • $begingroup$
    "The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
    $endgroup$
    – Simply Beautiful Art
    2 hours ago








2




2




$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
2 hours ago




$begingroup$
You can't simply plug 0 into parts. Take $limlimits_{xto0}frac xx$ for example. Plugging 0 into the numerator implies the limit should be 0, but clearly the limit is 1.
$endgroup$
– Simply Beautiful Art
2 hours ago












$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
2 hours ago






$begingroup$
@SimplyBeautifulArt But,in some of the questions we do replace 'x' by the constant values such that it simplifies to a non zero number.Why are we allowed to do that?
$endgroup$
– Navneet Kumar
2 hours ago














$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
2 hours ago




$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
2 hours ago












$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
2 hours ago




$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
2 hours ago












$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
2 hours ago




$begingroup$
"The only thing that I need to keep in mind is that the expression doesn't become indeterminate." This is vague, and also the reason we have clear limit rules to follow.
$endgroup$
– Simply Beautiful Art
2 hours ago










3 Answers
3






active

oldest

votes


















1












$begingroup$

When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
$$lim_{xrightarrow0}frac{2x}{3x}$$
, and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.



What I suggest is that you differentiate both the numerator and the denominator to get
$$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
Now differentiate again, and you get
$$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
    $endgroup$
    – Navneet Kumar
    2 hours ago










  • $begingroup$
    Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
    $endgroup$
    – Larry
    1 hour ago





















1












$begingroup$

Hint



Compose Taylor series
$$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
$$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write



    $$begin{align}
    frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
    &=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
    end{align}$$



    Therefore, we find that



    $$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$






    share|cite|improve this answer









    $endgroup$













      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%2f3080181%2fusing-partial-limit%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
      $$lim_{xrightarrow0}frac{2x}{3x}$$
      , and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.



      What I suggest is that you differentiate both the numerator and the denominator to get
      $$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
      Now differentiate again, and you get
      $$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
        $endgroup$
        – Navneet Kumar
        2 hours ago










      • $begingroup$
        Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
        $endgroup$
        – Larry
        1 hour ago


















      1












      $begingroup$

      When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
      $$lim_{xrightarrow0}frac{2x}{3x}$$
      , and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.



      What I suggest is that you differentiate both the numerator and the denominator to get
      $$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
      Now differentiate again, and you get
      $$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
        $endgroup$
        – Navneet Kumar
        2 hours ago










      • $begingroup$
        Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
        $endgroup$
        – Larry
        1 hour ago
















      1












      1








      1





      $begingroup$

      When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
      $$lim_{xrightarrow0}frac{2x}{3x}$$
      , and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.



      What I suggest is that you differentiate both the numerator and the denominator to get
      $$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
      Now differentiate again, and you get
      $$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$






      share|cite|improve this answer









      $endgroup$



      When both the numerator and denominator are zero, you have to differentiate both of them to obtain the correct limit. You can't straightly make the numerator $0$. Think about it, if you have
      $$lim_{xrightarrow0}frac{2x}{3x}$$
      , and you make the numerator zero and differentiate the denominator, you will get zero, which clearly isn't correct.



      What I suggest is that you differentiate both the numerator and the denominator to get
      $$lim_{xrightarrow0}frac{pisin(x)sin(pi/2cos(x))}{4x}$$
      Now differentiate again, and you get
      $$lim_{xrightarrow0}frac{pileft[cos(x)sin(pi/2cos(x))-pi/2sin(x)(cos(pi/2cos(x)))right]}{4}=frac{pi}{4}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 2 hours ago









      LarryLarry

      2,1162826




      2,1162826












      • $begingroup$
        Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
        $endgroup$
        – Navneet Kumar
        2 hours ago










      • $begingroup$
        Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
        $endgroup$
        – Larry
        1 hour ago




















      • $begingroup$
        Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
        $endgroup$
        – Navneet Kumar
        2 hours ago










      • $begingroup$
        Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
        $endgroup$
        – Larry
        1 hour ago


















      $begingroup$
      Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
      $endgroup$
      – Navneet Kumar
      2 hours ago




      $begingroup$
      Thanks So Much! Please go through this once artofproblemsolving.com/community/q2h1766346p11567701 I just want to know how to decide which 'x' are safe to replace by their constant values?
      $endgroup$
      – Navneet Kumar
      2 hours ago












      $begingroup$
      Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
      $endgroup$
      – Larry
      1 hour ago






      $begingroup$
      Nikolas did a pretty good job explaining when it is safe to replace the values. Yes, when the denominator becomes zero, you don't want to replace the value. In this question, the denominator is zero, so you can't just replace the value. You would instead do L'Hospital or some other manipulations as presented in other answers posted here.
      $endgroup$
      – Larry
      1 hour ago













      1












      $begingroup$

      Hint



      Compose Taylor series
      $$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
      $$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint



        Compose Taylor series
        $$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
        $$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint



          Compose Taylor series
          $$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
          $$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.






          share|cite|improve this answer









          $endgroup$



          Hint



          Compose Taylor series
          $$cos(x)=1-frac{x^2}{2}+Oleft(x^4right)$$
          $$cos left(frac{pi}{2} cos (x)right)=sin left(frac{pi }{4}x^2+Oleft(x^4right)right)$$ The next step is simple.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Claude LeiboviciClaude Leibovici

          120k1157132




          120k1157132























              1












              $begingroup$

              Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write



              $$begin{align}
              frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
              &=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
              end{align}$$



              Therefore, we find that



              $$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write



                $$begin{align}
                frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
                &=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
                end{align}$$



                Therefore, we find that



                $$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write



                  $$begin{align}
                  frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
                  &=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
                  end{align}$$



                  Therefore, we find that



                  $$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$






                  share|cite|improve this answer









                  $endgroup$



                  Using $frac{1-cos(x)}{2}=sin^2(x/2)$ and $lim_{tto 0}frac{sin(t)}{t}=1$, we can write



                  $$begin{align}
                  frac{cosleft(fracpi2 cos(x)right)}{x^2}&=frac{sinleft(pi sin^2(x/2)right)}{x^2}\\
                  &=fracpi4underbrace{left(frac{sin(x/2)}{x/2}right)^2}_{to1,,text{as},,xto0}underbrace{left(frac{sinleft(pi sin^2(x/2)right)}{pisin^2(x/2)}right)}_{to 1,,text{as},,xto0}\\
                  end{align}$$



                  Therefore, we find that



                  $$lim_{xto0}frac{cosleft(fracpi2 cos(x)right)}{x^2}=fracpi4$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Mark ViolaMark Viola

                  131k1275171




                  131k1275171






























                      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%2f3080181%2fusing-partial-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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more