Bài tập 42 trang 167 SGK Toán 11 NC

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

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

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - 2} \frac{{{x^3} + 8}}{{x + 2}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{x + 2}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \left( {{x^2} - 2x + 4} \right) = 12
\end{array}\)

c) \(\mathop {\lim }\limits_{x \to 9} \frac{{3 - \sqrt x }}{{9 - x}} = \mathop {\lim }\limits_{x \to 9} \frac{1}{{3 + \sqrt x }} = \frac{1}{6}\)

d)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{2 - \sqrt {4 - x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{4 - \left( {4 - x} \right)}}{{x\left( {2 + \sqrt {4 - x} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2 + \sqrt {4 - x} }} = \frac{1}{4}
\end{array}\)

e)

\(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^4} - {x^3} + 11}}{{2x - 7}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} - {x^2} + \frac{{11}}{x}}}{{2 - \frac{7}{x}}} =  + \infty \)

f) Với x < 0, ta có: 

\(\frac{{\sqrt {{x^4} + 4} }}{{x + 4}} = \frac{{{x^2}\sqrt {1 + \frac{4}{{{x^4}}}} }}{{x + 4}} = \frac{{x\sqrt {1 + \frac{4}{{{x^4}}}} }}{{1 + \frac{4}{x}}}\)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } x\sqrt {1 + \frac{4}{{{x^4}}}}  =  - \infty ,\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{4}{x}} \right) = 1 > 0\) 

nên \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^4} + 4} }}{{x + 4}} =  - \infty \)

-- Mod Toán 11 HỌC247