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
    1 hour 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
    1 hour ago












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










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












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










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









Larry

2,1162826




2,1162826










asked 2 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
    1 hour 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
    1 hour ago












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










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












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










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














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




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












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




$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
1 hour 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
1 hour 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
1 hour 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
    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
    50 mins 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
        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
        50 mins 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
        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
        50 mins 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 1 hour 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
        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
        50 mins 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
        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
        50 mins 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
      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
      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
      50 mins 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
      50 mins 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 1 hour 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 1 hour 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

                      Сан-Квентин

                      Image
                      Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но...
                      Read more

                      8-я гвардейская общевойсковая армия

                      Image
                      8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
                      Read more

                      Алькесар

                      Image
                      Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
                      Read more