Question

Đã trả lời bởi chuyên gia

Answer 1

Đặt $L = \lim_{x \to 0} (\frac{x}{\sqrt[7]{x + 1} . \sqrt{x + 4} - 2}) = \frac{a}{b}$ thì $\frac{1}{L} = \lim (\frac{\sqrt[7]{x + 1} . \sqrt{x + 4} - 2}{x}) = \frac{b}{a}$

Ta xét $\frac{b}{a} = \lim_{x \to 0} (\frac{\sqrt[7]{x + 1} . \sqrt{x + 4} - \sqrt{x + 4} + \sqrt{x + 4} - 2}{x})$

$= \lim_{x \to 0} (\frac{\sqrt[7]{x + 1} . \sqrt{x + 4} - \sqrt{x + 4}}{x}) + \lim_{x \to 0} (\frac{\sqrt{x + 4} - 2}{x}) = L_{1} + L_{2}$

Ta có $L_{1} = \lim_{x \to 0} [\frac{\sqrt{x + 4} (\sqrt[7]{x + 1} - 1)}{x}]$

Đặt $t = \sqrt[7]{x + 1}$ . Khi đó $\{\begin{cases} x = t^{7} - 1 \\ x \to 0 \Rightarrow t \to 1 \end{cases}$

$L_{1} = \lim_{t \to 1} \frac{\sqrt{t^{7} + 3} . (t - 1)}{t^{7} - 1} = \lim_{t \to 1} \frac{\sqrt{t^{7} + 3}}{(t^{6} + t^{5} + t^{4} + t^{3} + t^{2} + t + 1)} = \frac{2}{7}$

Xét $L_{2} = \lim_{x \to 0} (\frac{\sqrt{x + 4} - 2}{x}) = \lim_{x \to 0} \frac{(\sqrt{x + 4} - 2) (\sqrt{x + 4} + 2)}{x (\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2} = \frac{1}{4}$

Vậy

\frac{b}{a} = \frac{2}{7} + \frac{1}{4} = \frac{15}{28} \Rightarrow \frac{b}{a} = \frac{28}{15} \Rightarrow \lim_{x \to 0} (\frac{x}{\sqrt[7]{x + 1} . \sqrt{x + 4} - 2}) = \frac{28}{15}

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Answer 2