Đáp án:
\(\lim\limits_{x\to 1}\dfrac{\sqrt{5-x^3} - \sqrt[3]{x^2 + 7}}{x^2 - 1}= - \dfrac{11}{24}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to 1}\dfrac{\sqrt{5-x^3} - \sqrt[3]{x^2 + 7}}{x^2 - 1}\\
= \lim\limits_{x\to 1}\dfrac{\sqrt{5-x^3}-2 +2 - \sqrt[3]{x^2 + 7}}{x^2 - 1}\\
=\lim\limits_{x\to 1}\dfrac{\sqrt{5-x^3} -2}{x^2 - 1} + \lim\limits_{x\to 1}\dfrac{2 - \sqrt[3]{x^2 + 7}}{x^2 - 1}\\
= \lim\limits_{x\to 1}\dfrac{\left(\sqrt{5-x^3} -2\right)\left(\sqrt{5-x^3} + 2\right)}{(x^2 - 1)\left(\sqrt{5-x^3} + 2\right)} + \lim\limits_{x\to 1}\dfrac{\left(2 - \sqrt[3]{x^2 + 7}\right)\left[4 + 2\sqrt[3]{x^2 + 7} + \sqrt[3]{(x^2 + 7)^2}\right]}{(x^2 - 1)\left[4 + 2\sqrt[3]{x^2 + 7} + \sqrt[3]{(x^2 + 7)^2}\right]}\\
= \lim\limits_{x\to 1}\dfrac{1 - x^3}{(x^2 - 1)\left(\sqrt{5-x^3} + 2\right)}+\lim\limits_{x\to 1}\dfrac{1 - x^2}{(x^2 - 1)\left[4 + 2\sqrt[3]{x^2 + 7} + \sqrt[3]{(x^2 + 7)^2}\right]}\\
= \lim\limits_{x\to 1}\dfrac{(1-x)(1 + x + x^2)}{(x-1)(x+1)\left(\sqrt{5-x^3} + 2\right)}-\lim\limits_{x\to 1}\dfrac{x^2-1}{(x^2 - 1)\left[4 + 2\sqrt[3]{x^2 + 7} + \sqrt[3]{(x^2 + 7)^2}\right]}\\
= -\lim\limits_{x\to 1}\dfrac{1 + x + x^2}{(x+1)\left(\sqrt{5-x^3} + 2\right)}-\lim\limits_{x\to 1}\dfrac{1}{4 + 2\sqrt[3]{x^2 + 7} + \sqrt[3]{(x^2 + 7)^2}}\\
= -\dfrac{1 + 1 + 1^2}{(1+1)\left(\sqrt{5-1^3} + 2\right)}-\dfrac{1}{4 + 2\sqrt[3]{1^2 + 7} + \sqrt[3]{(1^2 + 7)^2}}\\
= - \dfrac{11}{24}
\end{array}\)
