Find $a, binmathbb{R}$ so that $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$ is finite.












1












$begingroup$



Find $a, binmathbb{R}$ so that
$$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.




I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?






calculus limits







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  • 1




    $begingroup$
    try finding for which value of $a$ and $b$, $1$ is a double root of $ax^6+bx^5+1$.
    $endgroup$
    – J.F
    Jan 20 at 14:49










  • $begingroup$
    I thought about that, but I have no idea how to do it.
    $endgroup$
    – JustAnAmateur
    Jan 20 at 14:51
















1












$begingroup$



Find $a, binmathbb{R}$ so that
$$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.




I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    try finding for which value of $a$ and $b$, $1$ is a double root of $ax^6+bx^5+1$.
    $endgroup$
    – J.F
    Jan 20 at 14:49










  • $begingroup$
    I thought about that, but I have no idea how to do it.
    $endgroup$
    – JustAnAmateur
    Jan 20 at 14:51














1












1








1





$begingroup$



Find $a, binmathbb{R}$ so that
$$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.




I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?






calculus limits







share|cite|improve this question











$endgroup$





Find $a, binmathbb{R}$ so that
$$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.




I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?






calculus limits




calculus limits






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share|cite|improve this question













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edited Jan 20 at 17:04









Blue

48.5k870154




48.5k870154










asked Jan 20 at 14:47









JustAnAmateurJustAnAmateur

876




876








  • 1




    $begingroup$
    try finding for which value of $a$ and $b$, $1$ is a double root of $ax^6+bx^5+1$.
    $endgroup$
    – J.F
    Jan 20 at 14:49










  • $begingroup$
    I thought about that, but I have no idea how to do it.
    $endgroup$
    – JustAnAmateur
    Jan 20 at 14:51














  • 1




    $begingroup$
    try finding for which value of $a$ and $b$, $1$ is a double root of $ax^6+bx^5+1$.
    $endgroup$
    – J.F
    Jan 20 at 14:49










  • $begingroup$
    I thought about that, but I have no idea how to do it.
    $endgroup$
    – JustAnAmateur
    Jan 20 at 14:51








1




1




$begingroup$
try finding for which value of $a$ and $b$, $1$ is a double root of $ax^6+bx^5+1$.
$endgroup$
– J.F
Jan 20 at 14:49




$begingroup$
try finding for which value of $a$ and $b$, $1$ is a double root of $ax^6+bx^5+1$.
$endgroup$
– J.F
Jan 20 at 14:49












$begingroup$
I thought about that, but I have no idea how to do it.
$endgroup$
– JustAnAmateur
Jan 20 at 14:51




$begingroup$
I thought about that, but I have no idea how to do it.
$endgroup$
– JustAnAmateur
Jan 20 at 14:51










3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint:



Use the following lemma:




Lemma: $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.




Proof: If $ax^6+bx^5+1 = p(x)cdot (x-1)^2$ then $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure



(1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$frac{a+b+1}{0}$")



(2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then
$$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = lim_{xrightarrow 1} frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$frac{p(1)}{0}$").



So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
      $endgroup$
      – GReyes
      Jan 20 at 17:49



















    0












    $begingroup$

    The remainder of the division of $ax^6+bx^5+1$ by $(x-1)^2$ must be $rx+s$, for some $r$ and $s$. Thus $ax^6+bx^5+1=(x-1)^2q(x)+rx+s$ for some polynomial $q(x)$. Evaluating at $1$ we obtain
    $$
    a+b+1=r+s tag{1}
    $$
    If we consider the derivatives, $6ax^5+5bx^4=2(x-1)q(x)+(x-1)^2q(x)+r$. Evaluating again at $1$ we get
    $$
    6a+5b=r tag{2}
    $$



    Therefore
    $$
    frac{ax^6+bx^5+1}{(x-1)^2}=q(x)+frac{rx+s}{(x-1)^2}
    $$
    What's now the condition in order that the limit for $xto1$ to exist finite?



    Hint: write $rx+s=r(x-1)+r+s$, so the fraction is
    $$
    frac{r}{(x-1)}+frac{r+s}{(x-1)^2}
    $$



    Note that conditions $(1)$ and $(2)$ have been obtained in a purely algebraic way (derivatives of polynomials don't need limits), so this doesn't use l'Hôpital.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint:



      Use the following lemma:




      Lemma: $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.




      Proof: If $ax^6+bx^5+1 = p(x)cdot (x-1)^2$ then $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure



      (1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$frac{a+b+1}{0}$")



      (2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then
      $$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = lim_{xrightarrow 1} frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$frac{p(1)}{0}$").



      So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint:



        Use the following lemma:




        Lemma: $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.




        Proof: If $ax^6+bx^5+1 = p(x)cdot (x-1)^2$ then $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure



        (1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$frac{a+b+1}{0}$")



        (2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then
        $$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = lim_{xrightarrow 1} frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$frac{p(1)}{0}$").



        So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint:



          Use the following lemma:




          Lemma: $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.




          Proof: If $ax^6+bx^5+1 = p(x)cdot (x-1)^2$ then $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure



          (1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$frac{a+b+1}{0}$")



          (2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then
          $$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = lim_{xrightarrow 1} frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$frac{p(1)}{0}$").



          So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.






          share|cite|improve this answer









          $endgroup$



          Hint:



          Use the following lemma:




          Lemma: $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.




          Proof: If $ax^6+bx^5+1 = p(x)cdot (x-1)^2$ then $lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure



          (1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$frac{a+b+1}{0}$")



          (2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then
          $$lim_{xto 1} frac{ax^6+bx^5+1}{(x-1)^2} = lim_{xrightarrow 1} frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$frac{p(1)}{0}$").



          So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 14:53









          YankoYanko

          7,0081629




          7,0081629























              1












              $begingroup$

              First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
                $endgroup$
                – GReyes
                Jan 20 at 17:49
















              1












              $begingroup$

              First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
                $endgroup$
                – GReyes
                Jan 20 at 17:49














              1












              1








              1





              $begingroup$

              First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.






              share|cite|improve this answer









              $endgroup$



              First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 20 at 17:02









              GReyesGReyes

              1,45515




              1,45515












              • $begingroup$
                If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
                $endgroup$
                – GReyes
                Jan 20 at 17:49


















              • $begingroup$
                If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
                $endgroup$
                – GReyes
                Jan 20 at 17:49
















              $begingroup$
              If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
              $endgroup$
              – GReyes
              Jan 20 at 17:49




              $begingroup$
              If the limit of a fraction is finite and the denominator has limit zero, the numerator should have limit zero as well. This is obvious, and the only fact I used above.
              $endgroup$
              – GReyes
              Jan 20 at 17:49











              0












              $begingroup$

              The remainder of the division of $ax^6+bx^5+1$ by $(x-1)^2$ must be $rx+s$, for some $r$ and $s$. Thus $ax^6+bx^5+1=(x-1)^2q(x)+rx+s$ for some polynomial $q(x)$. Evaluating at $1$ we obtain
              $$
              a+b+1=r+s tag{1}
              $$
              If we consider the derivatives, $6ax^5+5bx^4=2(x-1)q(x)+(x-1)^2q(x)+r$. Evaluating again at $1$ we get
              $$
              6a+5b=r tag{2}
              $$



              Therefore
              $$
              frac{ax^6+bx^5+1}{(x-1)^2}=q(x)+frac{rx+s}{(x-1)^2}
              $$
              What's now the condition in order that the limit for $xto1$ to exist finite?



              Hint: write $rx+s=r(x-1)+r+s$, so the fraction is
              $$
              frac{r}{(x-1)}+frac{r+s}{(x-1)^2}
              $$



              Note that conditions $(1)$ and $(2)$ have been obtained in a purely algebraic way (derivatives of polynomials don't need limits), so this doesn't use l'Hôpital.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The remainder of the division of $ax^6+bx^5+1$ by $(x-1)^2$ must be $rx+s$, for some $r$ and $s$. Thus $ax^6+bx^5+1=(x-1)^2q(x)+rx+s$ for some polynomial $q(x)$. Evaluating at $1$ we obtain
                $$
                a+b+1=r+s tag{1}
                $$
                If we consider the derivatives, $6ax^5+5bx^4=2(x-1)q(x)+(x-1)^2q(x)+r$. Evaluating again at $1$ we get
                $$
                6a+5b=r tag{2}
                $$



                Therefore
                $$
                frac{ax^6+bx^5+1}{(x-1)^2}=q(x)+frac{rx+s}{(x-1)^2}
                $$
                What's now the condition in order that the limit for $xto1$ to exist finite?



                Hint: write $rx+s=r(x-1)+r+s$, so the fraction is
                $$
                frac{r}{(x-1)}+frac{r+s}{(x-1)^2}
                $$



                Note that conditions $(1)$ and $(2)$ have been obtained in a purely algebraic way (derivatives of polynomials don't need limits), so this doesn't use l'Hôpital.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The remainder of the division of $ax^6+bx^5+1$ by $(x-1)^2$ must be $rx+s$, for some $r$ and $s$. Thus $ax^6+bx^5+1=(x-1)^2q(x)+rx+s$ for some polynomial $q(x)$. Evaluating at $1$ we obtain
                  $$
                  a+b+1=r+s tag{1}
                  $$
                  If we consider the derivatives, $6ax^5+5bx^4=2(x-1)q(x)+(x-1)^2q(x)+r$. Evaluating again at $1$ we get
                  $$
                  6a+5b=r tag{2}
                  $$



                  Therefore
                  $$
                  frac{ax^6+bx^5+1}{(x-1)^2}=q(x)+frac{rx+s}{(x-1)^2}
                  $$
                  What's now the condition in order that the limit for $xto1$ to exist finite?



                  Hint: write $rx+s=r(x-1)+r+s$, so the fraction is
                  $$
                  frac{r}{(x-1)}+frac{r+s}{(x-1)^2}
                  $$



                  Note that conditions $(1)$ and $(2)$ have been obtained in a purely algebraic way (derivatives of polynomials don't need limits), so this doesn't use l'Hôpital.






                  share|cite|improve this answer









                  $endgroup$



                  The remainder of the division of $ax^6+bx^5+1$ by $(x-1)^2$ must be $rx+s$, for some $r$ and $s$. Thus $ax^6+bx^5+1=(x-1)^2q(x)+rx+s$ for some polynomial $q(x)$. Evaluating at $1$ we obtain
                  $$
                  a+b+1=r+s tag{1}
                  $$
                  If we consider the derivatives, $6ax^5+5bx^4=2(x-1)q(x)+(x-1)^2q(x)+r$. Evaluating again at $1$ we get
                  $$
                  6a+5b=r tag{2}
                  $$



                  Therefore
                  $$
                  frac{ax^6+bx^5+1}{(x-1)^2}=q(x)+frac{rx+s}{(x-1)^2}
                  $$
                  What's now the condition in order that the limit for $xto1$ to exist finite?



                  Hint: write $rx+s=r(x-1)+r+s$, so the fraction is
                  $$
                  frac{r}{(x-1)}+frac{r+s}{(x-1)^2}
                  $$



                  Note that conditions $(1)$ and $(2)$ have been obtained in a purely algebraic way (derivatives of polynomials don't need limits), so this doesn't use l'Hôpital.







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                  answered Jan 20 at 17:29









                  egregegreg

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