Bài tập 4.22 trang 165 SBT Toán 11

a) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 2x - 3}}{{x - 1}} = \frac{{{3^2} - 2.3 - 3}}{{3 - 1}} = 0\)

b) \(\mathop {\lim }\limits_{x \to  - 2} h\left( x \right) = \mathop {\lim }\limits_{x \to  - 2} \frac{{2{x^3} + 15}}{{{{\left( {x + 2} \right)}^2}}} =  - \infty \)

Vì \(\mathop {\lim }\limits_{x \to  - 2} \left( {2{x^3} + 15} \right) =  - 1 < 0\) và \(\mathop {\lim }\limits_{x \to  - 3} {\left( {x + 2} \right)^2} = 0\)

c) 

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

d) \(\mathop {\lim }\limits_{x \to  - {2^ + }} h\left( x \right) = \mathop {\lim }\limits_{x \to  - {2^ + }} \frac{{x - 15}}{{x + 2}} =  - \infty \)

Vì \(\mathop {\lim }\limits_{x \to  - {2^ + }} (x - 15) =  - 17 < 0\) và \(\mathop {\lim }\limits_{x \to  - {2^ + }} \left( {x + 2} \right) = 0,x + 2 > 0,\forall x >  - 2\)

\(\mathop {\lim }\limits_{x \to  - {2^ + }} \left( {x + 2} \right) = 0,\mathop {\lim }\limits_{x \to  - {2^ - }} h\left( x \right) = \mathop {\lim }\limits_{x \to  - {2^ - }} \frac{{x - 15}}{{x + 2}} =  + \infty ,x + 2 > 0,\forall x >  - 2\)

Vì \(\mathop {\lim }\limits_{x \to  - {2^ - }} (x - 15) =  - 17 < 0\) và \(\mathop {\lim }\limits_{x \to  - {2^ - }} \left( {x + 2} \right) = 0,x + 2 < 0,\forall x <  - 2\)

-- Mod Toán 11 HỌC247