Đáp án:
$e)\text{2 TCN: }y=\pm \dfrac{1}{3}\\ j)\text{2 TCN: }y=\pm \dfrac{3}{2}.$
Giải thích các bước giải:
$e)y=\dfrac{x+1}{3|x|+2}\\ \displaystyle \lim_{x \to +\infty}\dfrac{x+1}{3|x|+2} =\displaystyle \lim_{x \to +\infty}\dfrac{x+1}{3x+2} =\displaystyle \lim_{x \to +\infty}\dfrac{1+\dfrac{1}{x}}{3+\dfrac{2}{x}} =\dfrac{1}{3}\\ \displaystyle \lim_{x \to -\infty}\dfrac{x+1}{3|x|+2} =\displaystyle \lim_{x \to -\infty}\dfrac{x+1}{-3x+2} =\displaystyle \lim_{x \to -\infty}\dfrac{1+\dfrac{1}{x}}{-3+\dfrac{2}{x}} =-\dfrac{1}{3}\\ \Rightarrow \text{TCN: }y=\pm \dfrac{1}{3}\\ j)y=\dfrac{\sqrt{9x^2-6x+1}}{2x+1} \ \ \ \ D=\mathbb{R} \setminus \left\{-\dfrac{1}{2}\right\}\\ =\dfrac{\sqrt{(3x-1)^2}}{2x+1}\\ =\dfrac{|3x-1|}{2x+1}\\ \displaystyle \lim_{x \to +\infty} \dfrac{|3x-1|}{2x+1} =\displaystyle \lim_{x \to +\infty} \dfrac{3x-1}{2x+1} =\displaystyle \lim_{x \to +\infty} \dfrac{3-\dfrac{1}{x}}{2+\dfrac{1}{x}} = \dfrac{3}{2}\\ \displaystyle \lim_{x \to -\infty} \dfrac{|3x-1|}{2x+1} =\displaystyle \lim_{x \to -\infty} \dfrac{-3x+1}{2x+1} =\displaystyle \lim_{x \to -\infty} \dfrac{-3+\dfrac{1}{x}}{2+\dfrac{1}{x}} =- \dfrac{3}{2}\\\Rightarrow \text{TCN: }y=\pm \dfrac{3}{2}.$
