Bài tập 45 trang 167 SGK Toán 11 NC

a)

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

b)

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

c)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ - }} \frac{{3 - x}}{{\sqrt {27 - {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{{{\left( {\sqrt {3 - x} } \right)}^2}}}{{\sqrt {\left( {3 - x} \right)\left( {{x^2} + 3x + 9} \right)} }}\\
 = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{\sqrt {3 - x} }}{{\sqrt {{x^2} + 3x + 9} }} = 0
\end{array}\)

d)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {{x^3} - 8} }}{{{x^2} - 2x}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} }}{{x\left( {x - 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{x}\sqrt {\frac{{{x^2} + 2x + 4}}{{x - 2}}}  =  + \infty 
\end{array}\)

(vì \(\mathop {\lim }\limits_{x \to {2^ + }} \sqrt {{x^2} + 2x + 4}  = 2\sqrt 3 ,\mathop {\lim }\limits_{x \to {2^ + }} x\sqrt {x - 2}  = 0\) và \(x\sqrt {x - 2}  > 0,\forall x > 2\)).

-- Mod Toán 11 HỌC247