Tìm các giới hạn sau:1) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{2{x^2} + 5x + 2}}{{{x^3} - 2x + 4}}\)2) \(\mathop {\lim

1) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{2{x^2} + 5x + 2}}{{{x^3} - 2x + 4}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x + 2} \right)\left( {2x + 1} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 2} \right)}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{2x + 1}}{{{x^2} - 2x + 2}} =  - \frac{3}{{10}}\)

2) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {{x^2} + 5}  - 3}}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{\left( {{x^2} - 3x + 2} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{x + 2}}{{\left( {x - 1} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}} = \frac{2}{3}
\end{array}\)

3) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} - 2x + 5}  + 3x - 1}}{{2x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {1 - \frac{2}{x} + \frac{5}{{{x^2}}}}  + 3x - 1}}{{2x + 1}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {1 - \frac{2}{x} + \frac{5}{{{x^2}}}}  + 3 - \frac{1}{x}}}{{2 + \frac{1}{x}}} = 1
\end{array}\)

4) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {4{x^2} - 3x + 1}  - 2x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{4{x^2} - 3x + 1 - 4{x^2}}}{{\sqrt {4{x^2} - 3x + 1}  + 2x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 3x + 1}}{{x\sqrt {4 - \frac{3}{x} + \frac{1}{{{x^2}}}}  + 2x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 3 + \frac{1}{x}}}{{\sqrt {4 - \frac{3}{x} + \frac{1}{{{x^2}}}}  + 2}} =  - \frac{3}{4}
\end{array}\)

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

Vì \(\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } x =  - \infty \\
\mathop {\lim }\limits_{x \to  + \infty } x\frac{{2 - \frac{3}{{{x^2}}} + \frac{5}{{{x^3}}}}}{{\frac{3}{{{x^2}}} - \frac{1}{x} - 2}} =  - 1
\end{array} \right.\)