giứp em bài tập 3 và 4 với ạ

Giải thích các bước giải:

\(\begin{array}{l}
3,\\
1,\\
\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} - 5x + 1}}{{{x^2} - 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{3 - \frac{5}{x} + \frac{1}{{{x^2}}}}}{{1 - \frac{2}{{{x^2}}}}} = \frac{{3 - 0 + 0}}{{1 - 0}} = \frac{3}{1} = 3\\
2,\\
\mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {x - 1} \right)}^2}{{\left( {7x + 2} \right)}^2}}}{{{{\left( {2x + 1} \right)}^4}}}\\
 = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2}}}.\frac{{{{\left( {7x + 2} \right)}^2}}}{{{x^2}}}:\frac{{{{\left( {2x + 1} \right)}^4}}}{{{x^4}}}} \right]\\
 = \mathop {\lim }\limits_{x \to \infty } \left[ {{{\left( {\frac{{x - 1}}{x}} \right)}^2}.{{\left( {\frac{{7x + 2}}{x}} \right)}^2}:{{\left( {\frac{{2x + 1}}{x}} \right)}^4}} \right]\\
 = \mathop {\lim }\limits_{x \to \infty } \left[ {{{\left( {1 - \frac{1}{x}} \right)}^2}.{{\left( {7 + \frac{2}{x}} \right)}^2}:{{\left( {2 + \frac{1}{x}} \right)}^4}} \right]\\
 = {1^2}{.7^2}:{2^4}\\
 = \frac{{49}}{{16}}
\end{array}\)

\(\begin{array}{l}
3,\\
\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {2{x^2} + 1} \right)\left( {5x + 3} \right)}}{{\left( {2{x^3} - 1} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\frac{{2{x^2} + 1}}{{{x^2}}}.\frac{{5x + 3}}{{{x^2}}}}}{{\frac{{2{x^3} - 1}}{{{x^3}}}.\frac{{x + 1}}{x}}}\\
 = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {2 + \frac{1}{{{x^2}}}} \right).\left( {\frac{5}{x} + \frac{3}{{{x^2}}}} \right)}}{{\left( {2 - \frac{1}{{{x^3}}}} \right).\left( {1 + \frac{1}{x}} \right)}}\\
 = \frac{{2.0}}{{2.1}} = 0\\
4,\\
\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - 4x}  - x} \right)\\
\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} - 4x}  - x} \right)\\
\mathop {\lim }\limits_{x \to  - \infty } \sqrt {{x^2} - 4x}  =  + \infty \\
\mathop {\lim }\limits_{x \to  - \infty } x =  - \infty \\
 \Rightarrow \mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} - 4x}  - x} \right) =  + \infty \\
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} - 4x}  - x} \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {\sqrt {{x^2} - 4x}  - x} \right)\left( {\sqrt {{x^2} - 4x}  + x} \right)}}{{\sqrt {{x^2} - 4x}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {{x^2} - 4x} \right) - {x^2}}}{{\sqrt {{x^2} - 4x}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 4x}}{{\sqrt {{x^2} - 4x}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 4}}{{\sqrt {1 - \frac{4}{x}}  + 1}} = \frac{{ - 4}}{{\sqrt 1  + 1}} =  - 2\\
4,\\
1,\\
\mathop {\lim }\limits_{x \to  + \infty } \left[ {x.\left( {\sqrt {{x^2} + 5}  - x} \right)} \right]\\
 = \mathop {\lim }\limits_{x \to  + \infty } \left[ {x.\frac{{\left( {{x^2} + 5} \right) - {x^2}}}{{\sqrt {{x^2} + 5}  + x}}} \right]\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{5x}}{{\sqrt {{x^2} + 5}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{5}{{\sqrt {1 + \frac{5}{{{x^2}}}}  + 1}}\\
 = \frac{5}{{\sqrt 1  + 1}} = \frac{5}{2}\\
2,\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} - x + 3}  + x} \right) =  + \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} - x + 3}  + x} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {{x^2} - x + 3} \right) - {x^2}}}{{\sqrt {{x^2} - x + 3}  - x}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x + 3}}{{\left| x \right|.\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}}  - x}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x + 3}}{{ - x.\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}}  - x}}\,\,\,\,\,\,\,\left( {x \to  - \infty  \Rightarrow x < 0 \Rightarrow \left| x \right| =  - x} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - 1 + \frac{3}{x}}}{{ - \sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}}  - 1}} = \frac{{ - 1}}{{ - \sqrt 1  - 1}} = \frac{1}{2}
\end{array}\)