Đáp án đúng: B
Giải chi tiết:ĐKXĐ: \(x \ge - 9,\,\,x \ne 0,x \ne - 1.\)
Ta có:
\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {x + 9} - 3}}{{\left( {{x^2} + x} \right)\left( {x + 10} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {\dfrac{1}{{{x^5}}} + \dfrac{9}{{{x^6}}}} - \dfrac{3}{{{x^3}}}}}{{\left( {1 + \dfrac{1}{x}} \right)\left( {1 + \dfrac{{10}}{x}} \right)}} = \dfrac{0}{1} = 0 \Rightarrow \) Đồ thị hàm số có TCN là \(y = 0.\)
\(\mathop {\lim }\limits_{x \to - {9^ + }} \dfrac{{\sqrt {x + 9} - 3}}{{\left( {{x^2} + x} \right)\left( {x + 10} \right)}} = \dfrac{{ - 3}}{{72}} = - \dfrac{1}{{24}}\)
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 9} - 3}}{{\left( {{x^2} + x} \right)\left( {x + 10} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{\left( {{x^2} + x} \right)\left( {x + 10} \right)\left( {\sqrt {x + 9} + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\left( {x + 1} \right)\left( {x + 10} \right)\left( {\sqrt {x + 9} + 3} \right)}} = \dfrac{1}{{60}}\end{array}\)
\(\mathop {\lim }\limits_{x \to - {1^ + }} \dfrac{{\sqrt {x + 9} - 3}}{{({x^2} + x)(x + 10)}} = + \infty \,\) , \(\mathop {\lim }\limits_{x \to - {1^ - }} \dfrac{{\sqrt {x + 9} - 3}}{{({x^2} + x)(x + 10)}} = - \infty \,\).
\( \Rightarrow \) Đồ thị hàm số có đúng 1 TCĐ là \(x = - 1\).
Chọn: B.
