Hiển nhiên là ĐTHS ko có TCĐ
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3x+4}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-x-1}{\sqrt{x^2+2x+3}+\sqrt{x^2+3x+4}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-1-\dfrac{1}{x}}{\sqrt{1+\dfrac{2}{x}+\dfrac{3}{x^2}}+\sqrt{1+\dfrac{3}{x}+\dfrac{4}{x^2}}}\)
\(=\dfrac{-1}{1+1}=-\dfrac{1}{2}\)
\(\Rightarrow y=-\dfrac{1}{2}\) là 1 TCN
\(\lim\limits_{x\rightarrow-\infty}y=\lim\limits_{x\rightarrow-\infty}\dfrac{-x-1}{\sqrt{x^2+2x+3}+\sqrt{x^2+3x+4}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-1-\dfrac{1}{x}}{-\sqrt{1+\dfrac{2}{x}+\dfrac{3}{x^2}}-\sqrt{1+\dfrac{3}{x}+\dfrac{4}{x^2}}}=\dfrac{-1}{-1-1}=\dfrac{1}{2}\)
\(\Rightarrow y=\dfrac{1}{2}\) là TCN
ĐTHS ko có TCĐ và có 2 TCN: \(y=\pm\dfrac{1}{2}\)
