Bài tập 35 trang 163 SGK Toán 11 NC

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

(vì \(\mathop {\lim }\limits_{x \to {2^ + }} \left( {2x + 1} \right) = 5,\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0\) và \(x - 2 > 0,\forall x > 2\)) 

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

(vì \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {2x + 1} \right) = 5,\mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = 0\) và \(x - 2 < 0,\forall x < 2\))

c) \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x - 1}}{{{x^2}}} =  - \infty \) 

(vì \(\mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) =  - 1 < 0,\mathop {\lim }\limits_{x \to 0} {x^2} = 0\) và \({x^2} > 0,\forall x \ne 0\))

d)

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

(vì \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 1} \right) = 3,\mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 4} \right) = 0\) và \({x^2} - 4 < 0,\forall x \in \left( { - 2;2} \right)\))

-- Mod Toán 11 HỌC247