Question

Tính giới hạn:

\(\lim\limits_{x->1}\dfrac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)

Answer 1

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-2+2-\sqrt[3]{x^2+7}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{5-x^3-4}{\sqrt{5-x^3}+2}+\dfrac{8-x^2-7}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{\left(x-1\right)\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1+x+x^2}{\sqrt{5-x^3}+2}+\dfrac{1+x}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{x+1}\)

\(=\dfrac{\dfrac{1+1+1}{\sqrt{5-1}+2}+\dfrac{1+1}{4+2\sqrt[3]{1+7}+\sqrt[3]{8^2}}}{1+1}\)

=19/40