Đồ thị hàm số \(y=\frac{x+3}{{{x}^{3}}-3x}\) có bao nhiêu đường tiêm cận?

Chọn B

Ta có:

\(\begin{array}{l} \mathop {\lim }\limits_{x \to \pm \infty } {\kern 1pt} y = \mathop {\lim }\limits_{x \to \pm \infty } {\kern 1pt} \frac{{x + 3}}{{{x^3} - 3x}}\\ = \mathop {\lim }\limits_{x \to \pm \infty } {\kern 1pt} \frac{{\frac{1}{{{x^2}}} + \frac{3}{{{x^3}}}}}{{1 - \frac{3}{{{x^2}}}}} = 0\\ \Rightarrow TCN:{\kern 1pt} {\kern 1pt} y = 0.\\ \mathop {\lim }\limits_{x \to {0^ + }} {\kern 1pt} y = - \infty ;{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathop {\lim }\limits_{x \to {0^ - }} {\kern 1pt} y = + \infty \\ \Rightarrow TCD:{\kern 1pt} {\kern 1pt} {\kern 1pt} x = 0.\\ \mathop {\lim }\limits_{x \to {{\left( {\sqrt 3 } \right)}^ + }} {\kern 1pt} y = + \infty ;{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathop {\lim }\limits_{x \to {{\left( {\sqrt 3 } \right)}^ - }} {\kern 1pt} y = - \infty \\ \Rightarrow TCD:{\kern 1pt} {\kern 1pt} {\kern 1pt} x = \sqrt 3 .\\ \mathop {\lim }\limits_{x \to {{\left( { - \sqrt 3 } \right)}^ + }} {\kern 1pt} y = - \infty ;{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathop {\lim }\limits_{x \to {{\left( { - \sqrt 3 } \right)}^ - }} {\kern 1pt} y = + \infty \\ \Rightarrow TCD:{\kern 1pt} {\kern 1pt} {\kern 1pt} x = - \sqrt 3 . \end{array}\)

Vậy đồ thị hàm số \(y=\frac{x+3}{{{x}^{3}}-3x}\) có \(4\) đường tiêm cận.