\(\lim\limits_{x\rightarrow1^+}\dfrac{2x-5}{x^2+x-2}=\lim\limits_{x\rightarrow1^+}\dfrac{2\cdot1-5}{1^2+1-2}=-\infty\)
=>Tiệm cận đứng là x=1
\(\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{2x-5}{x^2+x-2}=\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{2\cdot\left(-2\right)-5}{\left(-2\right)^2+\left(-2\right)-2}=-\infty\)
=>Tiệm cận đứng là x=-2
\(\lim\limits_{x\rightarrow+\infty}\dfrac{2x-5}{x^2+x-2}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2}{x}-\dfrac{5}{x^2}}{1+\dfrac{1}{x}-\dfrac{2}{x^2}}=0\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x-5}{x^2+x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}-\dfrac{5}{x^2}}{1+\dfrac{1}{x}-\dfrac{2}{x^2}}=0\)
=>\(\lim\limits_{x\rightarrow+\infty}\dfrac{2x-5}{x^2+x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{2x-5}{x^2+x-2}=0\)
=>Tiệm cận ngang là y=0
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