Question

\(lim_{x->\frac{+}{ }\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}\)

Answer 1

Giới hạn này tiến đến đâu vậy bạn? 2 trường hợp khác nhau đúng ko?

\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}=\lim\limits_{x\rightarrow+\infty}\frac{x\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=1\)

\(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}=\lim\limits_{x\rightarrow-\infty}\frac{\left|x\right|\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=\lim\limits_{x\rightarrow-\infty}\frac{-x\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=-1\)