Tính được các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} - 2x} \right) = 4\)

b) \(\mathop {\lim }\limits_{x \to - 2} \frac{{4{x^2} + 2x + 1}}{{x - 4}} = \frac{{13}}{6}\)

c) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - {x^2} + 2x - 24}}{{{x^2} - 9}} = \frac{{23}}{6}\)

d) \(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} - x - 14}}{{{x^2} - 7x - 18}} = \frac{9}{{11}}\)

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