Limiting Limits

$\small{\dfrac{x^{3}-10x^{2}-25x+250}{x^{4}-149x^{2}+4900}=\dfrac{(x-10)(x^2-25)}{(x^2-100)(x^2-49)}}$ $~~~~~~~~~~~~~~~~~~~~~~\small{=\dfrac{x^2-25}{(x+10)(x^2-49)}}$ $\color{plum}{\lim _{x\rightarrow 10}\dfrac{x^2-25}{(x+10)(x^2-49)}}=\dfrac{10^2-25}{(10+10)(10^2-49)}=\dfrac{5}{68}$ $\Large\color{forestgreen}{5+68=73}$

Since it has the Indeterminate Form $\dfrac{0}{0}$ , use L'Hospital's Rule

$\lim_ {x\rightarrow 10} \dfrac{x^3-10x^2-25x+250}{x^4-149x^2+4900}$

$=\lim_ {x\rightarrow 10} \dfrac{3x^2-20x-25}{4x^3-298x}$

$=\dfrac{3(10^2)-20(10)-25}{4(10^3)-298(10)}$

$=\dfrac{300-200-25}{4000-2980}$

$=\dfrac{75}{1020}$

$=\dfrac{5}{68}$

$=\dfrac{5}{68}$

Finally,

$a+b=5+68=$ $\boxed{73}$

Let $f(x) = x^3-10x^2-25x+250$ and $g(x) = x^4-149x^2+4900$ . Since $\begin{aligned} \lim_{x \to 10} \frac{f(x)}{g(x)} = \frac{0}{0} \end{aligned}$ , we can use L'Hôpital's rule .

$\begin{aligned} \lim_{x \to 10} \frac{f(x)}{g(x)} & = \lim_{x \to 10} \frac{f'(x)}{g'(x)} \\ & = \lim_{x \to 10} \frac{3x^2-20x-25} {4x^3-298x} \\ & = \frac{75}{1020} = \frac{5}{68} \\ & \\ \Rightarrow a + b & = 5+68 = \boxed{73} \end{aligned}$