Bài tập 28 trang 158 SGK Toán 11 NC

a) Với x > 0, ta có: \(\frac{{x + 2\sqrt x }}{{x - \sqrt x }} = \frac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}} = \frac{{\sqrt x  + 2}}{{\sqrt x  - 1}}\)

Do đó \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{x + 2\sqrt x }}{{x - \sqrt x }} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt x  + 2}}{{\sqrt x  - 1}} = \frac{2}{{ - 1}} =  - 2\)

b) Với x < 2, ta có: \(\frac{{4 - {x^2}}}{{\sqrt {2 - x} }} = \frac{{\left( {2 - x} \right)\left( {2 + x} \right)}}{{\sqrt {2 - x} }} = \left( {x + 2} \right)\sqrt {2 - x} \)

Do đó \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{4 - {x^2}}}{{\sqrt {2 - x} }} = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 2} \right)\sqrt {2 - x}  = 0\)

c) Với x > - 1, ta có:

\(\begin{array}{l}
\frac{{{x^2} + 3x + 2}}{{\sqrt {{x^5} + {x^4}} }} = \frac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{{x^2}\sqrt {x + 1} }}\\
 = \frac{{\sqrt {x + 1} \left( {x + 2} \right)}}{{{x^2}}}
\end{array}\)

Do đó \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{{x^2} + 3x + 2}}{{\sqrt {{x^5} + {x^4}} }} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{\sqrt {x + 1} \left( {x + 2} \right)}}{{{x^2}}} = 0\)

d) Với - 3 < x < 3, ta có: \(\frac{{\sqrt {{x^2} - 7x + 12} }}{{\sqrt {9 - {x^2}} }} = \frac{{\sqrt {\left( {3 - x} \right)\left( {4 - x} \right)} }}{{\sqrt {\left( {3 - x} \right)\left( {3 + x} \right)} }} = \frac{{\sqrt {4 - x} }}{{\sqrt {3 + x} }}\)

Do đó \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{\sqrt {{x^2} - 7x + 12} }}{{\sqrt {9 - {x^2}} }} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{\sqrt {4 - x} }}{{\sqrt {3 + x} }} = \frac{{\sqrt 6 }}{6}\)

-- Mod Toán 11 HỌC247