Limit of a quartic polynomial Let $f(x)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$. If $\displaystyle \lim_{x\to0}\left[1+\frac{f(x)}{x^2}\right]=3$, then $f(2)$ is equal to:
(1) $-8$
(2) $-4$
(3) $0$
(4) $4$
The answer is $0$. Can anyone explain the answer with steps?
 A: Hint
$$\lim_{x\rightarrow 0}\frac{f(x)}{x^2}=\lim_{x\rightarrow 0}\left(ax^2+bx+c+\frac{d}{x}+\frac{e}{x^2}\right)=2$$
that give us,
$$ d=e=0 \quad \text{and}\quad c=2$$
so, 
$$f(x)=ax^4+bx^3+2x^2 \rightarrow f'(x)=4ax^3+3bx^2+4x$$
and $f(x)$ has $x=1$ and $x=2$ as extreme values, so
$$f'(1)=f'(2)=0$$
Can you finish?
A: I can give you some hints. As far as I can tell, the real question is to find the polynomial, and then the answer for the multiple choice question will be found by plugging in $x=2$.
First, write your polynomial as a generic degree four polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$. Then the limit condition says that
$$\lim_{x\to0}\left[1+ax^2bx+c+\frac{d}{x}+\frac{e}{x^2}\right]=3,$$
which immediately implies that $d=0$ and $e=0$ (do you see why?), and then you can also find the value of the coefficient $c$ from this.
Now you have your polynomial as $f(x)=ax^4+bx^3+cx^2$, with a known value of $c$ here. Find the derivative of $f(x)$:
$$f'(x)=4ax^3+3bx^2+2cx=4ax\left(x^2+\frac{3b}{4a}x+\frac{2c}{4a}\right).$$
But since $x=1$ and $x=2$ are extreme points, they are roots of the derivative, and so the quadratic part in the parentheses above is equal to $(x-1)(x-2)=x^2-3x+2$. By equating the coefficients (remember that $c$ already has a numerical value), we'll find the values of $a$ and $b$.
