# Câu 45 trang 167 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a.

\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt {{x^2} + x} - \sqrt x } \over {{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} {x^2 \over {{x^2}\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}} \cr & = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}} = + \infty \cr}

b.  $$\mathop {\lim }\limits_{x \to {1^ - }} x{{\sqrt {1 - x} } \over {2\sqrt {1 - x} + 1 - x}} = \mathop {\lim }\limits_{x \to {1^ - }} {x \over {2 + \sqrt {1 - x} }} = {1 \over 2}$$

c.

\eqalign{ & \mathop {\lim }\limits_{x \to {3^ - }} {{3 - x} \over {\sqrt {27 - {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ - }} {{{{\left( {\sqrt {3 - x} } \right)}^2}} \over {\sqrt {\left( {3 - x} \right)\left( {{x^2} + 3x + 9} \right)} }} \cr & = \mathop {\lim }\limits_{x \to {3^ - }} {{\sqrt {3 - x} } \over {\sqrt {{x^2} + 3x + 9} }} = 0 \cr}

d.

\eqalign{ & \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {{x^3} - 8} } \over {{x^2} - 2x}} = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} } \over {x\left( {x - 2} \right)}} \cr & = \mathop {\lim }\limits_{x \to {2^ + }} {1 \over x}\sqrt {{{{x^2} + 2x + 4} \over {x - 2}}} = + \infty \cr}

Vì

\eqalign{ & \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {{x^2} + 2x + 4} = 2\sqrt 3 \cr & \mathop {\lim }\limits_{x \to {2^ + }} x\sqrt {x - 2} = 0;\,x\sqrt {x - 2} > 0\,\forall x > 2 \cr}