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
    43 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





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      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
        43 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
        43 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
        43 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
        43 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
      43 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
      43 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












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                  answered 1 hour ago









                  Mark ViolaMark Viola

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