Let P(x) be a polynomial of degree 4 , having extremum at $x=1,x=2$ and $\lim_{x\to 0}\frac{x^2+P(x)}{x^2}=2$ Then the value of P(2) 
Let $P(x)$ be a polynomial of degree 4 , having extremum at $x=1,x=2$ and $$\lim_{x\to 0}\frac{x^2+P(x)}{x^2}=2.$$ Then what is the value of $P(2)$?

I worked out the limit using L'Hospital got a relation in terms of second derivative of $P$; the other derivative relations are that first derivatives are zero at 1,2. 
How can we interpretate these derivative equations to find the function?
Any help is welcome.
 A: Write $P(x) = a + bx + cx^2 + dx^3 + ex^4$. Given the limit statement, we have
$$\lim_{x \to 0} ex^2 + dx + (c + 1) + \frac{b}{x} + \frac{a}{x^2} = 2$$
The only way that this is possible is to have $a = b = 0$ and $c + 1 = 2$; hence $c = 1$.
Furthermore, we have $P'(1) = 0 = P'(2)$, which will give you two more equations in the unknowns $d$ and $e$.

Since $P'(x) = 2 x + 3dx^2 + 4ex^3$, we can evaluate at $1$ and $2$ to find
$$2 + 3d + 4e = 0$$
and
$$4(2 + 12d + 64 e) = 0$$
A: I know this is a little late, but instead of formulating the equations, one may also write the function $f'(x)$ first, and then integrate it.
Assuming $f'(x) = (x-2)(x-1)(x-k)$ and evaluating the limit through L'Hospital's rule, we have $k=0$.
On integrating this we have some $g(x) + c = f(x)$, where $c=0$ because of the limit condition.
A: $$\lim_{x\rightarrow 0}\frac{p(x)}{x^2}=1$$
So for the existance of limit $p(0)=0$ and using D L hopital rule
$$\lim_{x\rightarrow 0}\frac{p'(x)}{2x}=1$$
Again for existance of limit $p'(0)=0$
Given  $p(x)$ be a polynomial of degree $4$ , having extremum at  $x=1,x=2$
So $$p'(x)=A x(x-1)(x-2)=A(x^3-3x^2+2x)$$
$$\int p'(x)dx = p(x)=A\int (x^3-3x^2+2x)dx=A\bigg[\frac{x^4}{4}-x^3+x^2\bigg]+\mathcal{C}$$
Put $x=0$ we have $\mathcal{C}=0$
And put $p'(x)$ into  $$\lim_{x\rightarrow 0}\frac{p'(x)}{2x}=1$$
We have $\displaystyle A=0.5$
So $$p(x)=\frac{1}{2}\bigg(\frac{x^4}{4}-x^3+x^2\bigg)$$
So $$p(2)=0$$
