Using Partial Limit












3












$begingroup$


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



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











$endgroup$








  • 3




    $begingroup$
    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
    Jan 20 at 4:49










  • $begingroup$
    @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
    Jan 20 at 4:57












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    Jan 20 at 4:58










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    Jan 20 at 5:01






  • 1




    $begingroup$
    "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
    Jan 20 at 5:06
















3












$begingroup$


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



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











$endgroup$








  • 3




    $begingroup$
    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
    Jan 20 at 4:49










  • $begingroup$
    @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
    Jan 20 at 4:57












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    Jan 20 at 4:58










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    Jan 20 at 5:01






  • 1




    $begingroup$
    "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
    Jan 20 at 5:06














3












3








3





$begingroup$


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



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 Jan 20 at 4:29









Larry

2,39131129




2,39131129










asked Jan 20 at 4:27









Navneet KumarNavneet Kumar

3981416




3981416








  • 3




    $begingroup$
    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
    Jan 20 at 4:49










  • $begingroup$
    @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
    Jan 20 at 4:57












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    Jan 20 at 4:58










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    Jan 20 at 5:01






  • 1




    $begingroup$
    "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
    Jan 20 at 5:06














  • 3




    $begingroup$
    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
    Jan 20 at 4:49










  • $begingroup$
    @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
    Jan 20 at 4:57












  • $begingroup$
    Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
    $endgroup$
    – Simply Beautiful Art
    Jan 20 at 4:58










  • $begingroup$
    @SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
    $endgroup$
    – Navneet Kumar
    Jan 20 at 5:01






  • 1




    $begingroup$
    "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
    Jan 20 at 5:06








3




3




$begingroup$
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
Jan 20 at 4:49




$begingroup$
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
Jan 20 at 4:49












$begingroup$
@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
Jan 20 at 4:57






$begingroup$
@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
Jan 20 at 4:57














$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58




$begingroup$
Limit rules and continuous functions satisfy $limlimits_{xto a}f(x)=f(a)$.
$endgroup$
– Simply Beautiful Art
Jan 20 at 4:58












$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01




$begingroup$
@SimplyBeautifulArt please go through this conversation once artofproblemsolving.com/community/q2h1766346p11567701
$endgroup$
– Navneet Kumar
Jan 20 at 5:01




1




1




$begingroup$
"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
Jan 20 at 5:06




$begingroup$
"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
Jan 20 at 5:06










3 Answers
3






active

oldest

votes


















2












$begingroup$

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
    Jan 20 at 5:03










  • $begingroup$
    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
    Jan 20 at 5:48





















2












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
      $endgroup$
      – Mark Viola
      Jan 30 at 5:07











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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
      Jan 20 at 5:03










    • $begingroup$
      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
      Jan 20 at 5:48


















    2












    $begingroup$

    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
      Jan 20 at 5:03










    • $begingroup$
      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
      Jan 20 at 5:48
















    2












    2








    2





    $begingroup$

    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 Jan 20 at 4:47









    LarryLarry

    2,39131129




    2,39131129












    • $begingroup$
      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
      Jan 20 at 5:03










    • $begingroup$
      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
      Jan 20 at 5:48




















    • $begingroup$
      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
      Jan 20 at 5:03










    • $begingroup$
      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
      Jan 20 at 5:48


















    $begingroup$
    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
    Jan 20 at 5:03




    $begingroup$
    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
    Jan 20 at 5:03












    $begingroup$
    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
    Jan 20 at 5:48






    $begingroup$
    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
    Jan 20 at 5:48













    2












    $begingroup$

    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









    $endgroup$


















      2












      $begingroup$

      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









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 Jan 20 at 4:43









        Claude LeiboviciClaude Leibovici

        121k1157133




        121k1157133























            2












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:07
















            2












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:07














            2












            2








            2





            $begingroup$

            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 Jan 20 at 5:19









            Mark ViolaMark Viola

            132k1275173




            132k1275173












            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:07


















            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:07
















            $begingroup$
            Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
            $endgroup$
            – Mark Viola
            Jan 30 at 5:07




            $begingroup$
            Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
            $endgroup$
            – Mark Viola
            Jan 30 at 5:07


















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