Question

tìm tiệm cận đứng, tiệm cận ngang, tiệm cận xiên

a) \(y=3x+2-\dfrac{2}{x-5}\)

b) \(y=\dfrac{1-\sqrt{x+1}}{x^2+2x}\)

c) \(y=\dfrac{\sqrt{x^2+x-1}}{x^2+2x}\)

Answer 1

a) \(y=3x+2-\dfrac{2}{x-5}\left(x\ne5\right)\)

\(\lim\limits_{x\rightarrow5}\left(3x+2-\dfrac{2}{x-5}\right)=\infty\) \(\Rightarrow TCĐ:x=5\)

\(\lim\limits_{x\rightarrow\infty}\left(3x+2-\dfrac{2}{x-5}\right)=\infty\) nên không có \(TCN\)

\(\lim\limits_{x\rightarrow-\infty}\left[y-\left(3x+2\right)\right]=\)\(\lim\limits_{x\rightarrow+\infty}\left[y-\left(3x+2\right)\right]=0\) \(\Rightarrow TCX:y=3x+2\)

Answer 2

b) \(y=\dfrac{1-\sqrt{x+1}}{x^2+2x}=\dfrac{1-\sqrt{x+1}}{x\left(x+2\right)}\left(x\ge-1;x\ne0;-2\right)\)

\(\lim\limits_{x\rightarrow0}\left(\dfrac{1-\sqrt{x+1}}{x\left(x+2\right)}\right)=\)\(\lim\limits_{x\rightarrow-2}\dfrac{1-\sqrt{x+1}}{x\left(x+2\right)}=\infty\) \(\Rightarrow TCĐ:x=0;x=-2\)

\(\lim\limits_{x\rightarrow\infty}\left(\dfrac{1-\sqrt{x+1}}{x\left(x+2\right)}\right)=0\) \(\Rightarrow TCN:y=0\)

Hàm số không có tiệm cận xiên vì bậc nhỏ chia bậc lớn

c) \(y=\dfrac{\sqrt{x^2+x-1}}{x^2+2x}=\dfrac{\sqrt{x^2+x-1}}{x\left(x+2\right)}\left(x\le\dfrac{-1-\sqrt{5}}{2}\cup x\ge\dfrac{-1+\sqrt{5}}{2};x\ne0;-2\right)\)

\(\lim\limits_{x\rightarrow0}\left(\dfrac{\sqrt{x^2+x-1}}{x\left(x+2\right)}\right)=\)\(\lim\limits_{x\rightarrow-2}\left(\dfrac{\sqrt{x^2+x-1}}{x\left(x+2\right)}\right)=\infty\) 

\(\Rightarrow TCĐ:x=0;x=-2\)

\(\lim\limits_{x\rightarrow\infty}\left(\dfrac{\sqrt{x^2+x-1}}{x\left(x+2\right)}\right)=\)\(\lim\limits_{x\rightarrow\infty}\dfrac{x\sqrt{1+\dfrac{1}{x}-\dfrac{1}{x^2}}}{x^2\left(1+\dfrac{2}{x}\right)}=0\)

\(\Rightarrow TCN:y=0\)

Hàm số không có tiệm cận xiên vì bậc nhỏ chia bậc lớn