Multivariable Limit for Some Function












0












$begingroup$


I am trying to compute the following limit and am trying to determine a way to compute



$$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}.$$



The equation "blows up" when $(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)$.



$$lim_{(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)} f(mu _1,mu _2,mu _3,mu_4)$$



I tried using L'Hopital's rule, but the problem is that it does not apply to multivariable calculus.



Any direction would be greatly appreciated...






calculus limits multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    It doesn't make sense for a $4$-tuple to converge to $1.$
    $endgroup$
    – zhw.
    Jan 9 at 23:42










  • $begingroup$
    Ok - I am unclear of the notation I should use, but each of $mu_i$ approaches 1
    $endgroup$
    – PiE
    Jan 9 at 23:44










  • $begingroup$
    What is $lambda?$
    $endgroup$
    – zhw.
    Jan 9 at 23:47










  • $begingroup$
    It's just a constant > 0 and a real number
    $endgroup$
    – PiE
    Jan 9 at 23:54












  • $begingroup$
    If $lambda>0$, then the only problem at $(1,1,1,1)$ is with $(mu_2-mu_1)$ in the denominator. So the limit can’t exist (even as an infinite limit) since it would be $pminfty$ depending on the sign of that difference during approach. And, worse, if $mu_2=0$ or $mu_3=0$ during approach, the limit could be $0$.
    $endgroup$
    – MPW
    Jan 9 at 23:59


















0












$begingroup$


I am trying to compute the following limit and am trying to determine a way to compute



$$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}.$$



The equation "blows up" when $(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)$.



$$lim_{(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)} f(mu _1,mu _2,mu _3,mu_4)$$



I tried using L'Hopital's rule, but the problem is that it does not apply to multivariable calculus.



Any direction would be greatly appreciated...






calculus limits multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    It doesn't make sense for a $4$-tuple to converge to $1.$
    $endgroup$
    – zhw.
    Jan 9 at 23:42










  • $begingroup$
    Ok - I am unclear of the notation I should use, but each of $mu_i$ approaches 1
    $endgroup$
    – PiE
    Jan 9 at 23:44










  • $begingroup$
    What is $lambda?$
    $endgroup$
    – zhw.
    Jan 9 at 23:47










  • $begingroup$
    It's just a constant > 0 and a real number
    $endgroup$
    – PiE
    Jan 9 at 23:54












  • $begingroup$
    If $lambda>0$, then the only problem at $(1,1,1,1)$ is with $(mu_2-mu_1)$ in the denominator. So the limit can’t exist (even as an infinite limit) since it would be $pminfty$ depending on the sign of that difference during approach. And, worse, if $mu_2=0$ or $mu_3=0$ during approach, the limit could be $0$.
    $endgroup$
    – MPW
    Jan 9 at 23:59
















0












0








0





$begingroup$


I am trying to compute the following limit and am trying to determine a way to compute



$$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}.$$



The equation "blows up" when $(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)$.



$$lim_{(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)} f(mu _1,mu _2,mu _3,mu_4)$$



I tried using L'Hopital's rule, but the problem is that it does not apply to multivariable calculus.



Any direction would be greatly appreciated...






calculus limits multivariable-calculus







share|cite|improve this question











$endgroup$




I am trying to compute the following limit and am trying to determine a way to compute



$$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}.$$



The equation "blows up" when $(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)$.



$$lim_{(mu _1,mu _2,mu _3,mu_4)rightarrow (1 ,1,1,1)} f(mu _1,mu _2,mu _3,mu_4)$$



I tried using L'Hopital's rule, but the problem is that it does not apply to multivariable calculus.



Any direction would be greatly appreciated...






calculus limits multivariable-calculus




calculus limits multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 23:45







PiE

















asked Jan 9 at 23:40









PiEPiE

610411




610411












  • $begingroup$
    It doesn't make sense for a $4$-tuple to converge to $1.$
    $endgroup$
    – zhw.
    Jan 9 at 23:42










  • $begingroup$
    Ok - I am unclear of the notation I should use, but each of $mu_i$ approaches 1
    $endgroup$
    – PiE
    Jan 9 at 23:44










  • $begingroup$
    What is $lambda?$
    $endgroup$
    – zhw.
    Jan 9 at 23:47










  • $begingroup$
    It's just a constant > 0 and a real number
    $endgroup$
    – PiE
    Jan 9 at 23:54












  • $begingroup$
    If $lambda>0$, then the only problem at $(1,1,1,1)$ is with $(mu_2-mu_1)$ in the denominator. So the limit can’t exist (even as an infinite limit) since it would be $pminfty$ depending on the sign of that difference during approach. And, worse, if $mu_2=0$ or $mu_3=0$ during approach, the limit could be $0$.
    $endgroup$
    – MPW
    Jan 9 at 23:59




















  • $begingroup$
    It doesn't make sense for a $4$-tuple to converge to $1.$
    $endgroup$
    – zhw.
    Jan 9 at 23:42










  • $begingroup$
    Ok - I am unclear of the notation I should use, but each of $mu_i$ approaches 1
    $endgroup$
    – PiE
    Jan 9 at 23:44










  • $begingroup$
    What is $lambda?$
    $endgroup$
    – zhw.
    Jan 9 at 23:47










  • $begingroup$
    It's just a constant > 0 and a real number
    $endgroup$
    – PiE
    Jan 9 at 23:54












  • $begingroup$
    If $lambda>0$, then the only problem at $(1,1,1,1)$ is with $(mu_2-mu_1)$ in the denominator. So the limit can’t exist (even as an infinite limit) since it would be $pminfty$ depending on the sign of that difference during approach. And, worse, if $mu_2=0$ or $mu_3=0$ during approach, the limit could be $0$.
    $endgroup$
    – MPW
    Jan 9 at 23:59


















$begingroup$
It doesn't make sense for a $4$-tuple to converge to $1.$
$endgroup$
– zhw.
Jan 9 at 23:42




$begingroup$
It doesn't make sense for a $4$-tuple to converge to $1.$
$endgroup$
– zhw.
Jan 9 at 23:42












$begingroup$
Ok - I am unclear of the notation I should use, but each of $mu_i$ approaches 1
$endgroup$
– PiE
Jan 9 at 23:44




$begingroup$
Ok - I am unclear of the notation I should use, but each of $mu_i$ approaches 1
$endgroup$
– PiE
Jan 9 at 23:44












$begingroup$
What is $lambda?$
$endgroup$
– zhw.
Jan 9 at 23:47




$begingroup$
What is $lambda?$
$endgroup$
– zhw.
Jan 9 at 23:47












$begingroup$
It's just a constant > 0 and a real number
$endgroup$
– PiE
Jan 9 at 23:54






$begingroup$
It's just a constant > 0 and a real number
$endgroup$
– PiE
Jan 9 at 23:54














$begingroup$
If $lambda>0$, then the only problem at $(1,1,1,1)$ is with $(mu_2-mu_1)$ in the denominator. So the limit can’t exist (even as an infinite limit) since it would be $pminfty$ depending on the sign of that difference during approach. And, worse, if $mu_2=0$ or $mu_3=0$ during approach, the limit could be $0$.
$endgroup$
– MPW
Jan 9 at 23:59






$begingroup$
If $lambda>0$, then the only problem at $(1,1,1,1)$ is with $(mu_2-mu_1)$ in the denominator. So the limit can’t exist (even as an infinite limit) since it would be $pminfty$ depending on the sign of that difference during approach. And, worse, if $mu_2=0$ or $mu_3=0$ during approach, the limit could be $0$.
$endgroup$
– MPW
Jan 9 at 23:59












1 Answer
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$begingroup$

$$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}$$ When the $mu_i to 1$, this reduces to
$$frac {lambda}{left(mu _2-mu _1right)(lambda +1) left(lambda ^2+lambda +1right) }$$ and then $cdots$






share|cite|improve this answer









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    $begingroup$

    $$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}$$ When the $mu_i to 1$, this reduces to
    $$frac {lambda}{left(mu _2-mu _1right)(lambda +1) left(lambda ^2+lambda +1right) }$$ and then $cdots$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}$$ When the $mu_i to 1$, this reduces to
      $$frac {lambda}{left(mu _2-mu _1right)(lambda +1) left(lambda ^2+lambda +1right) }$$ and then $cdots$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}$$ When the $mu_i to 1$, this reduces to
        $$frac {lambda}{left(mu _2-mu _1right)(lambda +1) left(lambda ^2+lambda +1right) }$$ and then $cdots$






        share|cite|improve this answer









        $endgroup$



        $$f(mu _1,mu _2,mu _3,mu_4)=frac{lambda mu _2 mu _3}{left(mu _2-mu _1right) left(lambda ^3+mu _3 left(2 lambda ^2+mu _2 left(2 lambda +mu _1right)right)right)}$$ When the $mu_i to 1$, this reduces to
        $$frac {lambda}{left(mu _2-mu _1right)(lambda +1) left(lambda ^2+lambda +1right) }$$ and then $cdots$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 6:43









        Claude LeiboviciClaude Leibovici

        123k1157134




        123k1157134






























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