Đáp án:
$b)\text{TCN: }y=\pm \dfrac{3}{2}\\ g) \text{TCN: }y=1.$
Giải thích các bước giải:
$b)y=\dfrac{3x}{\sqrt{4x^2+1}} \ \ \ \ D=\mathbb{R}\\ =\dfrac{3x}{2|x|\sqrt{1+\dfrac{1}{4x^2}}}\\ \displaystyle\lim_{x \to +\infty}\dfrac{3x}{2|x|\sqrt{1+\dfrac{1}{4x^2}}} =\displaystyle\lim_{x \to +\infty}\dfrac{3x}{2x\sqrt{1+\dfrac{1}{4x^2}}} =\displaystyle\lim_{x \to +\infty}\dfrac{3}{2\sqrt{1+\dfrac{1}{4x^2}}} =\dfrac{3}{2}\\ \displaystyle\lim_{x \to -\infty}\dfrac{3x}{2|x|\sqrt{1+\dfrac{1}{4x^2}}} =\displaystyle\lim_{x \to -\infty}\dfrac{3x}{-2x\sqrt{1+\dfrac{1}{4x^2}}} =\displaystyle\lim_{x \to -\infty}\dfrac{3}{-2\sqrt{1+\dfrac{1}{4x^2}}} =-\dfrac{3}{2}\\ \Rightarrow \text{TCN: }y=\pm \dfrac{3}{2}\\ g)y=1+\dfrac{\sqrt{x-2}}{x} \ \ \ \ D=[2;+\infty)\\ =\dfrac{x+\sqrt{x-2}}{x}\\ \displaystyle\lim_{x \to +\infty} \dfrac{x+\sqrt{x-2}}{x} = \displaystyle\lim_{x \to +\infty} \dfrac{1+\sqrt{\dfrac{1}{x}-\dfrac{2}{x^2}}}{1} =1\\ \Rightarrow \text{TCN: }y=1.$
