Find $p(2)$ of a polynomial $p$ 
Let $p$ be a polynomial of fourth degree having extremum at $x=1$ and $x=2$ and
  $\lim \limits_{x \to0}\left(1+\frac{p(x)}{x^2}\right)=2$. Then the value to $p(2)$ is?

This problem was in my book, I tried but I am not getting a clue as to how to begin.
 A: Note that the limit exists if 
$$\lim \limits_{x \to0} \frac{p(x)}{x^2}=1$$
and if $p(0)=0$, then by L'Hôpital
$$\lim \limits_{x \to0} \frac{p(x)}{x^2}=\lim \limits_{x \to0} \frac{p'(x)}{2x}=1$$
then $p'(0)=0$ and by L'Hôpital
$$\lim \limits_{x \to0} \frac{p'(x)}{2x}=\lim \limits_{x \to0} \frac{p''(x)}{2}=1\implies p''(0)=2$$
Moreover we know that $p'(1)=p'(2)=0$.
Then for $p(x)=ax^4+bx^3+cx^2+dx+e=0$ we deduce


*

*$p(0)=0 \implies e=0$

*$p'(0)=0 \implies d=0$

*$p''(0)=2 \implies 2c=2 \implies c=1 $


then $p(x)=ax^4+bx^3+x^2=0$ and now apply $p'(1)=0$ and $p'(2)=0$ that is


*

*$4a+3b+2=0$

*$32a+12b+4=0$


that is $a=\frac14$ and $b=-1$ then
$$p(x)=\frac14x^4-x^3+x^2$$
A: You can write $p(x)=A+Bx+Cx^2+Dx^3+Ex^4$; compute the derivative and evaluate it at $1$ and $2$, where it should be $0$. Also
$$
\lim_{x\to0}\frac{p(x)}{x^2}=1
$$
implies that $A=0$ and $B=0$ (why?), but also provides another condition.
In total you have five conditions that allow you to write the polynomial and compute its value at $2$.
A: Let $P(x)=Ax^4 +Bx^3+Cx^2+Dx+E.$  Since $\lim_{x\to 0} 1+[P(x)/x^2]=2,$ we must have $E=0$ and $D=0$ and $C=1.$  
So $P(x)=Ax^4+Bx^3+x^2.$
So $P'(x)=4A^3+3Bx^2+2x.$ 
Since $0=P'(1)=P'(2)$ we have $0=32A+12B+4=4A+3B+2,$ from which  $A$ and $B$ are found.
A: The finite limit implies $p=x^2+ax^3+bx^4$. The derivative $2x+3ax^2+4bx^3$ vanishes at $1$ and $2$, which gives simultaneous equations that obtain $a$ and $b$. Then $p(2)=4+8a+16b$ is trivial.
