Cho \(\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{x}{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - 2}}} \right) = \dfrac{a}{b}\) ($\dfrac{a}{b}$là phân số tối giản). Tính tổng \(L = a + b\).

Đặt \(L = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{x}{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - 2}}} \right) = \dfrac{a}{b}\) thì \(\dfrac{1}{L} = \lim \left( {\dfrac{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - 2}}{x}} \right) = \dfrac{b}{a}\).

Ta có

$\dfrac{b}{a} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - \sqrt {x + 4}  + \sqrt {x + 4}  - 2}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - \sqrt {x + 4} }}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {x + 4}  - 2}}{x}} \right)$

Xét ${L_1} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {x + 4} \left( {\sqrt[7]{{x + 1}} - 1} \right)}}{x}} \right)$. Đặt \(t = \sqrt[7]{{x + 1}}\). Khi đó :\(\left\{ \begin{array}{l}x = {t^7} - 1\\x \to 0 \Rightarrow t \to 1\end{array} \right.\)

\({L_1} = \mathop {\lim }\limits_{t \to 1} \dfrac{{\sqrt {{t^7} + 3} \left( {t - 1} \right)}}{{{t^7} - 1}} = \mathop {\lim }\limits_{t \to 1} \dfrac{{\sqrt {{t^7} + 3} }}{{\left( {{t^6} + {t^5} + {t^4} + {t^3} + {t^2} + t + 1} \right)}} = \dfrac{2}{7}\)

Xét ${L_2} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {x + 4}  - 2}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {x + 4}  - 2} \right)\left( {\sqrt {x + 4}  + 2} \right)}}{{x\left( {\sqrt {x + 4}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\sqrt {x + 4}  + 2}} = \dfrac{1}{4}$

Vậy \(\dfrac{b}{a} = \dfrac{2}{7} + \dfrac{1}{4} = \dfrac{{15}}{{28}}\)\( \Rightarrow a = 28,b = 15 \Rightarrow a + b = 43\) \( \Rightarrow a + b = 43\).