Tính giới hạn \(\mathop {\lim }\limits_{x \to + \infty } (1 - x)\left( {1 - {x^2}} \right)\left( {1 - {x^3}} \right)\).

Tìm được các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4 + x} - 2}}{{4x}} = \frac{1}{{16}}\);

b) \(\mathop {\lim }\limits_{x \to 2} \frac{{4 - {x^2}}}{{\sqrt {x + 7} - 3}} = - 24\)

c) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {2x + 5} - 3}}{{\sqrt {x + 2} - 2}} = \frac{4}{3}\)

d) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{x + 7}} - 2}}{{x - 1}} = \frac{1}{3}\)

Tìm được các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2} - 10x} \right) = + \infty \);

b) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{3{x^2} - 4x + 1}}{{2{x^2} + x + 1}} = \frac{3}{2}\)

c) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + x + 1} - 3x}}{{2 - 3x}} = \frac{5}{4}\)

d) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{8{x^3} + 3{x^2} + 1}} - x}}{{\sqrt {4{x^2} - x + 2} + 3x}} = 1\).