Tính các giới hạn sau: a) lim x-> - vô cùng x(x+1)(2x-1)/5x^3+x+7

Giải SBT Toán 11 Bài tập cuối chương 5 trang 87

Bài 5.46 trang 89 SBT Toán 11 Tập 1:Tính các giới hạn sau:

a) $\underset{x\to -\infty }{\lim }\frac{x\left(x+1\right)\left(2x-1\right)}{5x^3+x+7}$ ;

b) $\underset{x\to -\infty }{\lim }\left(x^3-1\right)\left(2-x^5\right)$ ;

c) $\underset{x\to +\infty }{\lim }\left(\sqrt[3]{x^3+x^2+1}-x\right)$ .

Lời giải:

a)$\underset{x\to -\infty }{\lim }\frac{x\left(x+1\right)\left(2x-1\right)}{5x^3+x+7}=\underset{x\to -\infty }{\lim }\frac{x^3\left(1+\frac{1}{x}\right)\left(2-\frac{1}{x}\right)}{x^3\left(5+\frac{1}{x^2}+\frac{7}{x^3}\right)}$

$=\underset{x\to -\infty }{\lim }\frac{\left(1+\frac{1}{x}\right)\left(2-\frac{1}{x}\right)}{5+\frac{1}{x^2}+\frac{7}{x^3}}=\frac{2}{5}$.

b)$\underset{x\to -\infty }{\lim }\left(x^3-1\right)\left(2-x^5\right)=\underset{x\to -\infty }{\lim }{x}^3\left(1-\frac{1}{x^3}\right)x^5\left(\frac{2}{x^5}-1\right)$

$=\underset{x\to -\infty }{\lim }{x}^8\left(1-\frac{1}{x^3}\right)\left(\frac{2}{x^5}-1\right)=-\infty$.

c)$\underset{x\to +\infty }{\lim }\left(\sqrt[3]{x^3+x^2+1}-x\right)$

$=\underset{x\to +\infty }{\lim }\frac{x^3+x^2+1-x^3}{\sqrt[3]{{\left(x^3+x^2+1\right)}^2}+x\sqrt[3]{x^3+x^2+1}+x^2}$

$=\underset{x\to +\infty }{\lim }\frac{x^2+1}{\sqrt[3]{{\left(x^3+x^2+1\right)}^2}+x\sqrt[3]{x^3+x^2+1}+x^2}$

$=\underset{x\to +\infty }{\lim }\frac{1+\frac{1}{x^2}}{\sqrt[3]{{\left(1+\frac{1}{x}+\frac{1}{x^3}\right)}^2}+\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^3}}+1}=\frac{1}{3}$