Tiệm cận đứng của đồ thị hàm số y= x-5/ x^2-x-6 là

A. $x=1.$

B. $y=-2;y=3.$

C. x=1

D. $x=-2;x=3.$

Trả lời

Chọn D

Tập xác định của hàm số là    $D=ℝ\backslash \left\{-2;3\right\}.$

Ta có $\underset{x\to {\left(-2\right)}^{-}}{\lim }y=\underset{x\to {\left(-2\right)}^{-}}{\lim }\frac{x-5}{x^2-x-6}=\underset{x\to {\left(-2\right)}^{-}}{\lim }\frac{x-5}{\left(x+2\right)\left(x-3\right)}=-\infty .$

Vì $\left(\begin{array}{l}\underset{x\to {\left(-2\right)}^{-}}{\lim }\left(x-5\right)=-7<0\\ \underset{x\to {\left(-2\right)}^{-}}{\lim }\left(x+2\right)\left(x-3\right)=0\\ \left(x+2\right)\left(x-3\right)>0,\forall x<-2\end{array}\right).$

Lại có $\underset{x\to {\left(-2\right)}^{+}}{\lim }y=\underset{x\to {\left(-2\right)}^{+}}{\lim }\frac{x-5}{x^2-x-6}=\underset{x\to {\left(-2\right)}^{+}}{\lim }\frac{x-5}{\left(x+2\right)\left(x-3\right)}=+\infty .$

Do $\left(\begin{array}{l}\underset{x\to {\left(-2\right)}^{+}}{\lim }\left(x-5\right)=-7<0\\ \underset{x\to {\left(-2\right)}^{+}}{\lim }\left(x+2\right)\left(x-3\right)=0\\ \left(x+2\right)\left(x-3\right)<0,\forall x>-2\end{array}\right).$

$\underset{x\to 3^{+}}{\lim }y=\underset{x\to 3^{+}}{\lim }\frac{x-5}{x^2-x-6}=\underset{x\to 3^{+}}{\lim }\frac{x-5}{\left(x+2\right)\left(x-3\right)}=-\infty .$

$\underset{x\to 3^{-}}{\lim }y=\underset{x\to 3^{-}}{\lim }\frac{x-5}{x^2-x-6}=\underset{x\to 3^{-}}{\lim }\frac{x-5}{\left(x+2\right)\left(x-3\right)}=+\infty .$