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

* Hướng dẫn giải

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \left( {x - 1} \right) = 2 - 1 = 1\)

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

c) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x + 3}  - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {\sqrt {{x^2} + x + 3}  - x} \right)\left( {\sqrt {{x^2} + x + 3}  + x} \right)}}{{\sqrt {{x^2} + x + 3}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} + x + 3 - {x^2}}}{{\sqrt {{x^2} + x + 3}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x + 3}}{{\sqrt {{x^2} + x + 3}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{3}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{3}{x}}  + 1}} = \frac{1}{2}
\end{array}\)