cj giải hộ e bài 2 này vs a

Đáp án:

6) 2

Giải thích các bước giải:

\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{x - 2}}\\
 = \mathop {\lim }\limits_{x \to 2} \left( {x - 3} \right)\\
 = 2 - 3 =  - 1\\
2)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {2x - 1} \right)}}{{x - 2}}\\
 = \mathop {\lim }\limits_{x \to 2} \left( {2x - 1} \right)\\
 = 2.2 - 1 = 3\\
3)\mathop {\lim }\limits_{x \to 2} \dfrac{{x + 2 - 4}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{\sqrt {x + 2}  + 2}}\\
 = \dfrac{1}{{2 + 2}} = \dfrac{1}{4}\\
4)\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{1 + \dfrac{1}{x}}}{{4 + \dfrac{3}{x}}} = \dfrac{1}{4}\\
5)\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{2 - \dfrac{1}{x}}}{{\dfrac{3}{x} - 1}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{2}{{ - 1}}\\
 =  - 2\\
6)\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {4 + \dfrac{1}{{{x^2}}}} }}{{1 + \dfrac{1}{x}}}\\
 = 2
\end{array}\)