For polynomial $g(x)$ satisfying $(g(a))^2+(g'(a))^2=0$, evaluate $lim_{xto...












1












$begingroup$



If $g(x)$ is a polynomial function and



$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate



$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$




Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$



means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$



So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$



with Squeeze Theorem, the limit must be equal to $1$.



But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$



Could some help me to explain it? Thanks.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you work only with real numbers?
    $endgroup$
    – Mindlack
    Jan 21 at 14:08










  • $begingroup$
    yes mindlack...
    $endgroup$
    – DXT
    Jan 21 at 14:11










  • $begingroup$
    Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
    $endgroup$
    – learner
    Jan 21 at 14:14










  • $begingroup$
    Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
    $endgroup$
    – Mindlack
    Jan 21 at 14:16






  • 1




    $begingroup$
    Learner: I am afraid your equation is wrong.
    $endgroup$
    – Mindlack
    Jan 21 at 14:19
















1












$begingroup$



If $g(x)$ is a polynomial function and



$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate



$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$




Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$



means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$



So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$



with Squeeze Theorem, the limit must be equal to $1$.



But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$



Could some help me to explain it? Thanks.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you work only with real numbers?
    $endgroup$
    – Mindlack
    Jan 21 at 14:08










  • $begingroup$
    yes mindlack...
    $endgroup$
    – DXT
    Jan 21 at 14:11










  • $begingroup$
    Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
    $endgroup$
    – learner
    Jan 21 at 14:14










  • $begingroup$
    Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
    $endgroup$
    – Mindlack
    Jan 21 at 14:16






  • 1




    $begingroup$
    Learner: I am afraid your equation is wrong.
    $endgroup$
    – Mindlack
    Jan 21 at 14:19














1












1








1





$begingroup$



If $g(x)$ is a polynomial function and



$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate



$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$




Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$



means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$



So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$



with Squeeze Theorem, the limit must be equal to $1$.



But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$



Could some help me to explain it? Thanks.






limits







share|cite|improve this question











$endgroup$





If $g(x)$ is a polynomial function and



$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate



$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$




Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$



means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$



So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$



with Squeeze Theorem, the limit must be equal to $1$.



But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$



Could some help me to explain it? Thanks.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 14:16









Blue

48.6k870154




48.6k870154










asked Jan 21 at 14:03









DXTDXT

5,7582731




5,7582731












  • $begingroup$
    Do you work only with real numbers?
    $endgroup$
    – Mindlack
    Jan 21 at 14:08










  • $begingroup$
    yes mindlack...
    $endgroup$
    – DXT
    Jan 21 at 14:11










  • $begingroup$
    Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
    $endgroup$
    – learner
    Jan 21 at 14:14










  • $begingroup$
    Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
    $endgroup$
    – Mindlack
    Jan 21 at 14:16






  • 1




    $begingroup$
    Learner: I am afraid your equation is wrong.
    $endgroup$
    – Mindlack
    Jan 21 at 14:19


















  • $begingroup$
    Do you work only with real numbers?
    $endgroup$
    – Mindlack
    Jan 21 at 14:08










  • $begingroup$
    yes mindlack...
    $endgroup$
    – DXT
    Jan 21 at 14:11










  • $begingroup$
    Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
    $endgroup$
    – learner
    Jan 21 at 14:14










  • $begingroup$
    Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
    $endgroup$
    – Mindlack
    Jan 21 at 14:16






  • 1




    $begingroup$
    Learner: I am afraid your equation is wrong.
    $endgroup$
    – Mindlack
    Jan 21 at 14:19
















$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08




$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08












$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11




$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11












$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14




$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14












$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16




$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16




1




1




$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19




$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19










1 Answer
1






active

oldest

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1












$begingroup$

$lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$



You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$



if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.



All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.



But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.



**I am new here, so excuse me if I got some mistakes.. trying my best :)






share|cite|improve this answer











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    $begingroup$

    $lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$



    You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$



    if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.



    All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.



    But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.



    **I am new here, so excuse me if I got some mistakes.. trying my best :)






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$



      You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$



      if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.



      All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.



      But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.



      **I am new here, so excuse me if I got some mistakes.. trying my best :)






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$



        You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$



        if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.



        All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.



        But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.



        **I am new here, so excuse me if I got some mistakes.. trying my best :)






        share|cite|improve this answer











        $endgroup$



        $lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$



        You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$



        if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.



        All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.



        But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.



        **I am new here, so excuse me if I got some mistakes.. trying my best :)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 21 at 14:38

























        answered Jan 21 at 14:27









        ShaqShaq

        3049




        3049






























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