1. Use the condition on extreme values

Since $f(x)$ is a polynomial of degree $4$ and has extreme values at $x=1$ and $x=2$, we must have $f'(1)=0 \quad \text{and} \quad f'(2)=0.$

Because $f'(x)$ is a cubic polynomial, it must contain the factors $(x-1)$ and $(x-2)$. So write $f'(x)=k(x-1)(x-2)(x-a)$ for some constants $k,a$.

2. Use the limit condition

Given $\lim_{x\to 0}\left[1+\frac{f(x)}{x^2}\right]=3.$ So, $\lim_{x\to 0}\frac{f(x)}{x^2}=2.$

For this limit to be finite, we need $f(0)=0 \quad \text{and} \quad f'(0)=0.$ Thus $x=0$ is a double root of $f(x)$, so $f(x)=x^2(ax^2+bx+c).$

Also, $\lim_{x\to 0}\frac{f(x)}{x^2}=c=2.$ Hence $f(x)=x^2(ax^2+bx+2).$

3. Differentiate $f(x)$

Expand first: $f(x)=ax^4+bx^3+2x^2.$ Then $f'(x)=4ax^3+3bx^2+4x.$

Since extrema occur at $x=1$ and $x=2$, $f'(1)=0, \qquad f'(2)=0.$

So,

• At $x=1$: $4a+3b+4=0 \quad ...(1)$
• At $x=2$: $32a+12b+8=0$ $8a+3b+2=0 \quad ...(2)$

4. Solve for $a$ and $b$

Subtract (1) from (2): $(8a+3b+2)-(4a+3b+4)=0$ $4a-2=0$ $a=\frac12.$

Substitute into (1): $4\left(\frac12\right)+3b+4=0$ $2+3b+4=0$ $3b=-6$ $b=-2.$

Therefore, $f(x)=\frac12 x^4-2x^3+2x^2.$

5. Find $f(2)$

$f(2)=\frac12(16)-2(8)+2(4)$ $=8-16+8=0.$

6. Conclusion

Thus, $\boxed{f(2)=0}.$ So the correct option is A.