Giới hạn \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 2} - 2}}{{x - 2}}\) bằng     

Tính giới hạn \(\mathop {\lim }\limits_{x \to 3} \frac{{2{x^2} - 5x - 3}}{{\sqrt {5x + 1} - 4}}\).

Cho hàm số f(x) = x² – 3x + 2.

a) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{x - 1}} = - 1\).

b) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^2} - 1}} = \frac{1}{4}\).

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^3} - {x^2} + x - 1}} > 0\).

d) Để \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = 2\) thì a + 3b = 1.

Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 3x + 1\;\;khi\;\;x < 0\\\sqrt {{x^2} + 1} \;\;\;\;\;\;khi\;\;x \ge 0\end{array} \right.\). Khi đó:

a) Giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = - 1\).

b) Giới hạn \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1\).

c) Giới hạn \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1\).

d) Giới hạn \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1\).

Cho hàm số \(f\left( x \right) = \frac{{\sqrt {x + 1} - 2}}{{x - 3}}\). Khi đó:

a) \(f\left( 8 \right) = - \frac{1}{5}\).

b) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \frac{1}{3}\).

c) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \frac{1}{6}\).

d) Biết \(\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = a,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 2} - x} \right) = b\). Khi đó 3a + 4b = 2.