Tính giới hạn \(B=\lim \limits_{x \rightarrow 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}\)
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\text { Ta có: } \mathrm{B}=\lim\limits _{x \rightarrow 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}=\lim \limits_{x \rightarrow 7} \frac{\frac{\sqrt{x+2}-3}{x-7}-\frac{\sqrt[3]{x+20}-3}{x-7}}{\frac{\sqrt[4]{x+9}-2}{x-7}}\)
\(\begin{array}{*{20}{l}} {{\rm{ Mà : }}\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{x - 7}} = \mathop {\lim }\limits_{x \to 7} \frac{1}{{\sqrt {x + 2} + 3}} = \frac{1}{6}}\\ {\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{x + 20}} - 3}}{{x - 7}} = \mathop {\lim }\limits_{x \to 7} \frac{1}{{{{(\sqrt[3]{{x + 20}})}^2} + 3\sqrt[3]{{x + 20}} + 9}} = \frac{1}{{27}}}\\ {\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[4]{{x + 9}} - 2}}{{x - 7}} = \mathop {\lim }\limits_{x \to 7} \frac{1}{{{{(\sqrt[4]{{x + 9}})}^3} + 2{{(\sqrt[4]{{x + 9}})}^2} + 4\sqrt[4]{{x + 9}} + 8}} = \frac{1}{{32}}}\\ {{\rm{ Vậy }}B = \frac{{\frac{1}{6} - \frac{1}{{27}}}}{{\frac{1}{{32}}}} = \frac{{112}}{{27}}.} \end{array}\)