Since $f$ is a polynomial of degree $4$ and has extreme values at $x=4$ and $x=5$, we must have $f'(4)=0 \quad \text{and} \quad f'(5)=0.$

Also, $f$ is degree $4$, so $f'$ is degree $3$. Hence we can write $f'(x)=k(x-4)(x-5)(x-a)$ for some constants $k\neq 0$ and $a$.

Now use the limit condition: $\lim_{x\to 0}\frac{f(x)}{x^2}=5.$ For this limit to exist and be finite, we need $f(0)=0.$ Also, since $f(x)=f(0)+f'(0)x+\cdots,$ for $\frac{f(x)}{x^2}$ to have a finite limit, we must also have $f'(0)=0.$ Then the limit equals $\frac{f''(0)}{2}=5 \implies f''(0)=10.$

Since $f'(0)=0$, and $f'(x)=k(x-4)(x-5)(x-a)$, substituting $x=0$ gives $0=f'(0)=k(-4)(-5)(-a)= -20ka.$ Because $k\neq 0$, we get $a=0.$ So, $f'(x)=k x(x-4)(x-5).$

Expand: $x(x-4)(x-5)=x(x^2-9x+20)=x^3-9x^2+20x.$ Thus $f'(x)=k(x^3-9x^2+20x).$

Differentiate to get $f''(x)$: $f''(x)=k(3x^2-18x+20).$ Hence $f''(0)=20k.$ But from step 3, $f''(0)=10$, so $20k=10 \implies k=\frac12.$

Therefore, $f'(x)=\frac12 x(x-4)(x-5)=\frac12(x^3-9x^2+20x).$ Integrate: