Đáp án:
$d)y=0\\ i)y=\pm \dfrac{2}{3}.$
Giải thích các bước giải:
$d)y=\dfrac{\sqrt{x-3}}{x-2} \ \ \ \ D=[3;+\infty)\\ \displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{x-3}}{x-2} =\displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{\dfrac{1}{x}-\dfrac{3}{x^2}}}{1-\dfrac{2}{x}} =0\\ \Rightarrow \text{TCN: }y=0\\ i)y=\dfrac{\sqrt{4x^2+1}}{\sqrt[3]{27x^3+x}} \ \ \ \ D=\mathbb{R}\\ \displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{4x^2+1}}{\sqrt[3]{27x^3+x}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{2|x|\sqrt{1+\dfrac{1}{4x^2}}}{3x\sqrt[3]{1+\dfrac{1}{27x^3}}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{2x\sqrt{1+\dfrac{1}{4x^2}}}{3x\sqrt[3]{1+\dfrac{1}{27x^3}}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{2\sqrt{1+\dfrac{1}{4x^2}}}{3\sqrt[3]{1+\dfrac{1}{27x^3}}}\\ =\dfrac{2}{3}\\ \displaystyle\lim_{x \to -\infty} \dfrac{\sqrt{4x^2+1}}{\sqrt[3]{27x^3+x}}\\ =\displaystyle\lim_{x \to -\infty} \dfrac{2|x|\sqrt{1+\dfrac{1}{4x^2}}}{3x\sqrt[3]{1+\dfrac{1}{27x^3}}}\\ =\displaystyle\lim_{x \to -\infty} \dfrac{-2x\sqrt{1+\dfrac{1}{4x^2}}}{3x\sqrt[3]{1+\dfrac{1}{27x^3}}}\\ =\displaystyle\lim_{x \to -\infty} \dfrac{-2\sqrt{1+\dfrac{1}{4x^2}}}{3\sqrt[3]{1+\dfrac{1}{27x^3}}}\\ =-\dfrac{2}{3}\\ \Rightarrow \text{TCN: }y=\pm \dfrac{2}{3}.$
