Bài tập 43 trang 167 SGK Toán 11 NC

Bài tập 43 trang 167 SGK Toán 11 NC

a)

\(\begin{array}{l}
\frac{{{x^3} + 3\sqrt 3 }}{{3 - {x^2}}} = \frac{{\left( {x + \sqrt 3 } \right)\left( {{x^2} - x\sqrt 3  + 3} \right)}}{{\left( {x + \sqrt 3 } \right)\left( {\sqrt 3  - x} \right)}}\\
 = \frac{{{x^2} - x\sqrt 3  + 3}}{{\sqrt 3  - x}}
\end{array}\)

Do đó:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \sqrt 3 } \frac{{{x^3} + 3\sqrt 3 }}{{3 - {x^2}}} = \mathop {\lim }\limits_{x \to  - \sqrt 3 } \frac{{{x^2} - x\sqrt 3  + 3}}{{\sqrt 3  - x}}\\
 = \frac{9}{{2\sqrt 3 }} = \frac{{3\sqrt 3 }}{2}
\end{array}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{{x^2} - 4x}} = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{x\left( {x - 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 4} \frac{1}{{x\left( {\sqrt x  + 2} \right)}} = \frac{1}{{16}}
\end{array}\)

c)

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

d)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + x + 1}  - 1}}{{3x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + x + 1 - 1}}{{3x\left( {\sqrt {{x^2} + x + 1}  + 1} \right)}}\\
 = \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{x + 1}}{{\sqrt {{x^2} + x + 1}  + 1}} = \frac{1}{6}
\end{array}\)

-- Mod Toán 11